In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0m away. After a running start, he leaps at a velocity of 5.0 m/s at an angle of 15degrees with respect to the flat roof. Will he make it to the other roof, which is 2.5m shorter than the building he jumps from?

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wornoutwomanC · Accepted Answer

Consider the vertical and horizobntal motions separately,  Given, x=4.0 m, initial velocity is 5.0 m/s at anangle of 15∘ with the horizontal, y=-2.5m.  The vx=vcos⁡15=5.0×0.9659=4.829 ms  vy=vsin⁡15=5.0×0.2588=1.294 ms  To find out the time flight, consider the vertical motion;  acceleration is- g=−9.8 ms2.  From S=ut+(12)at2;−2.5=5.0×0.2588t−4.9t2  =−2.5=1.294t−4.9t2  4.9t2−1.294t−2.5=0  t=+1.294±50.2942×4.9=1.294+7.1199.8  8.4129.8=0.858 seconds  The horizontal distance covered, x=(ucos⁡θ)t  =5.0×0.9659×0.858=4.11 m  As this distance is &amp;gt; the separation betwen thebuildings, the man will just make it.

Jeffrey Jordon · Accepted Answer

Explanation :  It is given that,  Distance between two building is 4 m.  Velocity of stuntman is 5 m/s.  He will follow a projectile motion.  We know that the horizontal component of velocity is given by :  vx=vcos⁡θ  and v=dt  So, t=dvcos⁡θ  t=45cos⁡15  t=0.833 s…(1)  Now, from second equation of motion :  y=ut−12gt2  since, for vertical component vy=vsin⁡θ  y=5×sin⁡(15)×0.833 s−12×9.8 m/s2×(0.833)2  y=2.35 s  So, he will make it to the other roof as 2.35 m is less than 2.50 m. He is able to cross the cliff without any injury.

xleb123 · Accepted Answer

Answer:If the stuntman&#039;s calculated horizontal distance is greater than or equal to 6.5m (4.0m + 2.5m), he will make it to the other roof. Otherwise, he will fall short.Explanation:The horizontal component of the stuntman&#039;s velocity can be calculated using the equation:Vx=V·cos(θ)where Vx is the horizontal component of velocity, V is the initial velocity of 5.0 m/s, and θ is the angle of 15 degrees.Substituting the given values, we have:Vx=5.0·cos(15∘)The vertical component of the stuntman&#039;s velocity can be calculated using the equation:Vy=V·sin(θ)where Vy is the vertical component of velocity, V is the initial velocity of 5.0 m/s, and θ is the angle of 15 degrees.Substituting the given values, we have:Vy=5.0·sin(15∘)The time of flight can be calculated using the equation:t=2·Vygwhere t is the time of flight, Vy is the vertical component of velocity, and g is the acceleration due to gravity (approximately 9.8m/s2).Substituting the given values, we have:t=2·(5.0·sin(15∘))9.8Now, we can calculate the horizontal distance traveled by the stuntman using the equation:D=Vx·twhere D is the horizontal distance, Vx is the horizontal component of velocity, and t is the time of flight.Substituting the previously calculated values, we have:D=(5.0·cos(15∘))·(2·(5.0·sin(15∘))9.8)Finally, we can compare the calculated horizontal distance to the actual distance of 4.0m to determine if the stuntman will make it to the other roof. Additionally, we need to consider that the other roof is 2.5m shorter than the building the stuntman jumps from. Therefore, if the calculated horizontal distance minus 2.5m is greater than or equal to 4.0m, the stuntman will make it to the other roof.

Jazz Frenia · Accepted Answer

To solve this problem, we can break it down into horizontal and vertical components. Let&#039;s first calculate the time it takes for the stuntman to reach the other building:The horizontal component of the velocity (vx) is given by:vx=vcos(θ)where v is the initial velocity (5.0 m/s) and θ is the angle of the jump (15 degrees). Substituting the values, we have:vx=5.0cos(15∘)Next, we can calculate the time it takes to reach the other building (t) using the horizontal distance (dx) and horizontal velocity (vx):dx=vx·twhere dx is the horizontal distance (4.0 m). Rearranging the equation, we have:t=dxvxSubstituting the values, we have:t=4.05.0cos(15∘)Now, let&#039;s calculate the vertical distance the stuntman jumps (dy):The vertical component of the velocity (vy) is given by:vy=vsin(θ)Substituting the values, we have:vy=5.0sin(15∘)Using the formula for vertical distance, we have:dy=vy·t+12gt2where g is the acceleration due to gravity (approximately 9.8 m/s^2). Substituting the values, we have:dy=(5.0sin(15∘))·(4.05.0cos(15∘))+12·9.8(4.05.0cos(15∘))2Finally, we need to check if the vertical distance dy is greater than or equal to the height difference between the buildings (2.5 m). If dy≥2.5, the stuntman will make it to the other roof.Let&#039;s calculate the values and check the condition:vx=5.0cos(15∘)≈4.844t=4.05.0cos(15∘)≈0.882vy=5.0sin(15∘)≈1.294dy=(5.0sin(15∘))·(4.05.0cos(15∘))+12·9.8(4.05.0cos(15∘))2≈1.524Since dy≈1.524 is greater than 2.5, the stuntman will make it to the other roof.Therefore, the stuntman will successfully jump to the other building in the action movie scene.

fudzisako · Accepted Answer

For the horizontal motion:x=vinitial·t·cos(θ)For the vertical motion:y=vinitial·t·sin(θ)−12gt2where:- vinitial is the initial velocity of the stuntman (5.0 m/s),- θ is the angle of the jump (15 degrees),- g is the acceleration due to gravity (approximately 9.8 m/s^2),- t is the time of flight.Since the stuntman jumps from the top of one building to the top of another building, the height of the buildings does not affect the horizontal motion. Therefore, we only need to consider the vertical motion.Let&#039;s solve for the time of flight first:y=vinitial·t·sin(θ)−12gt2Substituting the known values:2.5=5.0·t·sin(15)−12·9.8·t2Now, let&#039;s solve this quadratic equation for t.

In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0m away. After a running start, he leaps

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