A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal. (a) determine the time taken by the projectile to hitthe point P at ground level. (b) determine the range X of the projectile as measured from the base of the cliff at the instant just before the projectile hits point P. Find (x) the horizontal and vertical components of its velocity and (d) the magnitude of the velocity and (e) the angle made by the velocity vector with the horizontal (f) Find the maximum height above the cliff top reached by the projectile.

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Ezra Herbert · Accepted Answer

a)Find the time until projectile hits the ground  Use Y components  Yf=Yi+ViT+12gT2  Yf=−125m  Yi=0  Vi=65.0sin⁡37 make sure your calculator is set to degrees  g=−9.8 ms2  Solve for T  b)To find range use X coordinates  Xf=Xi+ViT+12aT2  Xi=0  Vi=65.0cos⁡37  a=0 this is always the case in projectile motion  Plug T from a) and find Xf  c)Remember there is no air friction on this problem, so accelaration is zero. That means velocity stays the same on the X coordinate.  As for the Y coordinate Vf=Vi+aT  You know T, a=-g, and Vi  The direction is down or course.  d)The magnitude is found by y2+x2 you know the drill  e) I think this is θ=arccos⁡ (x/magnitude found in d)  f) Vf=Vi+at use Y components  Vf=0 at the top  we know Vi and a=-g  find this T and use in the position formula to get Yf

Jeffrey Jordon · Accepted Answer

The height of the cliff above ground level, y0=125 m  The initial speed of the projectile, u = 65.0 m/s  The angle of elevation of the projectile, θ=37∘  (a) The time for the projectile to hit the ground, t, is given as follows;  y=y0+u⋅sin⁡(θ)⋅t−(12)⋅g⋅t2  g=9.81 m/s2  At ground level, y = 0, we get;  0=125+65⋅sin⁡(37)⋅t−4.905⋅t2  ²t=−65⋅sin⁡(37)±65⋅sin⁡(37)²−4×(−4.905)×125(2×(−4.905)  ∴t≈−2.45 or t≈10.42  The time it takes the projectile to hit the ground,t≈10.42 seconds  (b) The range of the projectile X=u⋅cos⁡(θ)×t  ∴X=65.0×cos⁡(37∘)×10.42≈540.92  The range of the projectile from the base of the cliff, X≈540.92 meters  (c) The horizontal component of the velocity, ux=u×cos⁡(θ)  ∴ux=65.0×cos⁡(37∘)≈51.91  The horizontal component of the velocity, ux≈51.91m/s  The vertical component of the velocity, uy=u×sin⁡(θ)  ∴uy=65.0×sin⁡(37∘)≈39.12  The vertical component of the velocity, uy≈39.12m/s  (d) The magnitude of the velocity, |u|, is the given speed  = 65.0 m/s  |u|=(uy2+ux2)  ∴|u|=(39.122+51.912)=65  |u| = 65.0 m/s  (e) The angle made by the velocity vector with the horizontal = The angle of elevation with which the projectile is launched = 37∘  (f) The maximum height above the cliff top reached by the projectile, ymax of cliff, is given as follows;  ymax of cliff=(u×sin⁡(θ))22⋅g  ∴ymax of cliff=65.0×sin⁡(37∘)22×9.81≈77.99  ymax of cliff≈77.99 meters

Nick Camelot · Accepted Answer

Step 1:(a) To determine the time taken by the projectile to hit point P at ground level, we can analyze the vertical motion. The initial vertical velocity is given by viy=visin(θ), where vi=65.0 m/s is the initial speed and θ=37∘ is the launch angle.The time taken to reach the maximum height can be found using the equation vfy=viy+gt, where vfy=0 m/s since the projectile reaches its maximum height. Solving for t gives us:0=viy+gtt=−viygThe time of flight, which is the total time taken by the projectile to reach the ground, can be found using the equation h=viyt+12gt2, where h=125 m is the height of the cliff. Substituting the expression for t gives us:125=viy(−viyg)+12g(−viyg)2Simplifying the equation:125=−viy2g+viy22g125=viy22gviy2=250gviy=250gSubstituting the known values:viy=250×9.8 m/sNow, we can find the total time of flight by substituting the expression for viy:t=−viygt=−250×9.89.8 sTherefore, the time taken by the projectile to hit point P at ground level is t=−250×9.89.8 s.Step 2:(b) To determine the range X of the projectile as measured from the base of the cliff at the instant just before the projectile hits point P, we can analyze the horizontal motion. The horizontal velocity remains constant throughout the motion and is given by vix=vicos(θ).The range X can be found using the equation X=vix×t, where vix=vicos(θ) and t is the time of flight obtained in part (a).Therefore, the range X of the projectile is X=vix×t=(65.0m/s×cos(37∘))×−250×9.89.8.Step 3:(x) The horizontal and vertical components of the velocity are given by vx=vicos(θ) and vy=visin(θ) respectively. Substituting the known values:vx=65.0m/s×cos(37∘)vy=65.0m/s×sin(37∘)Step 4:(d) The magnitude of the velocity is given by the Pythagorean theorem: v=vx2+vy2. Substituting the known values:v=(65.0m/s×cos(37∘))2+(65.0m/s×sin(37∘))2Step 5:(e) The angle made by the velocity vector with the horizontal is given by θv=arctan(vyvx). Substituting the known values:θv=arctan(65.0m/s×sin(37∘)65.0m/s×cos(37∘))Step 6:(f) The maximum height above the cliff top reached by the projectile can be found using the equation hmax=h+viy×t+12gt2, where hmax is the maximum height.hmax=125+250×9.8×(−250×9.89.8)+12×9.8×(−250×9.89.8)2

Mr Solver · Accepted Answer

(a) The vertical motion equation can be used to calculate how long it will take the projectile to arrive at point P at ground level: y=y0+v0yt−12gt2, where y0 is the initial vertical position, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.Given: y0=125m, v0y=v0sin(θ), v0=65.0m/s, θ=37∘, and g=9.8m/s2.Substituting the values into the equation, we have:0=125m+(65.0m/s)sin(37∘)t−12(9.8m/s2)t2.To find the time t, we need to solve this quadratic equation for t.(b) The range X of the projectile can be determined using the horizontal motion equation: X=v0xt, where v0x is the initial horizontal velocity.Given: v0x=v0cos(θ).Substituting the values, we have:X=(65.0m/s)cos(37∘)t.To find the range X, we need to substitute the value of t obtained from part (a) into this equation.(x) The horizontal and vertical components of the velocity can be calculated using the equations:vx=v0cos(θ) and vy=v0sin(θ).Substituting the given values, we get:vx=(65.0m/s)cos(37∘) and vy=(65.0m/s)sin(37∘).(d) The magnitude of the velocity v can be determined using the Pythagorean theorem:v=vx2+vy2.Substituting the values, we have:v=((65.0m/s)cos(37∘))2+((65.0m/s)sin(37∘))2.(e) The angle made by the velocity vector with the horizontal can be calculated using the equation:θv=arctan(vyvx).Substituting the values, we get:θv=arctan((65.0m/s)sin(37∘)(65.0m/s)cos(37∘)).(f) The maximum height above the cliff top reached by the projectile can be calculated using the formula:H=y0+v0y22g.Substituting the values, we have:H=125m+((65.0m/s)sin(37∘))22(9.8m/s2).

Eliza Beth13 · Accepted Answer

To solve the given problem, we&#039;ll break it down into smaller parts. Let&#039;s start with part (a).(a) Determining the time taken by the projectile to hit point P at ground level:Let&#039;s denote the initial speed as v0, the angle with the horizontal as θ, the time taken as t, and the acceleration due to gravity as g.Using the given information, we have:Initial vertical velocity, v0y=v0sin(θ),Vertical displacement, y=−125 m (negative because it&#039;s downward),Vertical acceleration, ay=−g (negative because it&#039;s downward).We can use the equation of motion to relate the variables:y=v0yt+12ayt2.Substituting the values:−125=(v0sin(θ))t−12gt2.Simplifying the equation, we get a quadratic equation in terms of t:−12gt2+(v0sin(θ))t−125=0.We can solve this quadratic equation to find the value of t.(b) Determining the range X of the projectile:The range is the horizontal distance covered by the projectile. We can calculate it using the equation:X=v0cos(θ)·t.We already have the value of t from part (a), so we can substitute it to find the range X.Now, let&#039;s move on to the other parts of the problem.(x) Determining the horizontal and vertical components of the velocity:The horizontal component of velocity, vx, can be calculated using:vx=v0cos(θ).The vertical component of velocity, vy, can be calculated using:vy=v0sin(θ).(d) Determining the magnitude of the velocity:The magnitude of the velocity, v, can be calculated using the Pythagorean theorem:v=vx2+vy2.(e) Determining the angle made by the velocity vector with the horizontal:The angle, α, made by the velocity vector with the horizontal can be calculated using:α=arctan(vyvx).(f) Finding the maximum height above the cliff top reached by the projectile:To find the maximum height, we need to calculate the vertical displacement at the highest point of the projectile&#039;s trajectory. We can use the equation:ymax=v0y2/(2ay).

A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal

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