Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.

Question

Bukvald5z · Accepted Answer

Step 1 By neglecting the resistance of the air, both stones will experience an acceleration only on the vertical axis which is the acceleration of the gravity, and the velocity in the horizontal axis is constant. Therefore, the following equation describes the motion of the stones in the vertical axis: y=v0yt+12at2 since both stones are thrown horizontally, the vertical initial velocity is zero, so: y=12at2 the first stone will make a vertical displacement magnitude equals the height of the first building H1 but with negative sign, during a time of flight t1, so: −H1=12at12 the second stone will make a verticaldisplacement magnitude equals the height of the second building H2 but with negative sign, during a time of flight t2, so: H2=12at22 Step 2 The distance in the horizontal axis is given by: x=v0xt t=xv0x (1) Both stones are thrown horizontally with the same velocity (say vx), so for the first stone this equation will become: t1=x1vx and for the second stone is, t2=x2vx hence, the height of the two building will become: H1=−12at12 H2=−12at22 H1=−12a(x1vx)2 H2=−12a(x2vx)2 but the first stone lands twice as far from the base of the base of the building which means x1=2x2, so: H1=−12a(2x2vx)2 H2=−12a(x2vx)2 clearly H1&amp;gt;H2, so the ratio, so the height of the taller building to the height of the shorter building is: H1H2=−12a(2x2vx)2−12a(x2vx)2

Papilys3q · Accepted Answer

Distance from the base of the building to place where stone landing: l=vt where v is velocity, t is time. Time can be found from: h=gt22 where h is height of the building t=2hg Therefore: l1=v2h1g,l2=v2h1g l1l2=h1h2=2 h1h2=4 Answer: 4

karton · Accepted Answer

Answer:4 times tallerExplanation:The horizontal distance travelled by a projectile isd=vxt(1)where vx is the horizontal velocity (which is constant) and t is the total time of the fall.The total time of the fall can be found by analyzing the vertical motion of the projectile: the vertical position at time t is given byy=h−12gt2where h is the initial height, g=9.8m/s2 is the acceleration due to gravity, t is the time.The total time of the fall is the time t at which y=0 (the projectile reaches the ground), so0=h−12gt2t=2hgSubstituting into (1),d=vx2hgThis can be re-arranged asd2=vx2h2gh=2gd2vx2so we see that the initial height depends on the square of the horizontal distance travelled, d.In this problem, one stands lands twice as far as the other stone does, sod&#039; = 2dthis means that the height of the taller builing is h′=2g(2d)2vx2=4(2gd2vx2)=4hso the taller building is 4 times taller than the smaller one.

Stones are thrown horizontally with the same velocity from the tops of

Answered question

Answer & Explanation

New Questions in Other