How tall, h, is the bulding?

Question

A 0.98kg rock is projected from the edge of the top of a building with an initial velocity of 12.4m/s at an angle 60degree above the horizontal . due to gravity, the rock strikes the ground at a horizontal distance of 17m from the base of the building.
Assume: the ground is level and that the side of thr bulding is verticle. the asseleration of gravity is 9.8m/s2

Answer 1

1. what are the vertical & horizontal components of the initial velocity?

vertical 12.4sin60 = 10.7m/s
horizontal 12.4cos60 = 6.2 m/s

2. assuming constant horizontal velocity (and the givens), how long is the rock airborne?

17m / 6.2m/s = 2.74s

3. how long until the rock returns to its initial height (i.e. the building's height)?

Review: the kinematic equations

d = vit + at²/2 ; vf² = vi² + 2ad ; vf = vi + at; d = t(vi+fv)/2

Using equation 3, 0 = 10.7m/s + t(9.8m/s²)
t = (10.7m/s) / 9.8m/s² = 1.09s

So it goes up for 1.09s before starting to fall, and it will take 1.09s to reach top of the building height again, for a total airtime of 2.18s. When it reaches building height it will be travelling towards the ground at 10.7m/s and it will have (2.74s-2.18s) = 0.56s left to fall. How far will it fall (i.e. what is the building's height)?

Using equation 1: d = vit + at²/2

d = 10.7m/s x 0.56 + 0.56²(9.8m/s²)/2 = 7.53m

Answer 2

1.-First, descompose the velocity in an horizontal component, and a vertical component.
2.-Then: you know the rock strikes the ground at 17m (horizontal), so you can get the time it takes a rock, with a initial velocity given by your initial horizontal velocity component, to reach 17m (in other words, solve another problem, getting the time, given the distance and the initial velocity).

3.-The time you got from step 2, is the time it takes the rock to go up, and then fall, in the vertical dimension (since is the time the rock strikes the ground). You have to divide the problem in the following steps:
3.a)Get the time it takes the rock to go up to the top of its trajectory. Then multiply it for 2, and you get the time it takes the rock to go up, and down, till the level of the top of the building. The distance traveled by the rock in this interval, doesn`t belong the the building height`s. Get also the final velocity, that`s the velocity when the rock is falling, and going downward.

3.b) Rest the time you got in (3.a) [the previous point] to the value you got in step 2, so you can get the time the rock is falling in front of the building wall. Now, with the velocity you got in (3.a) [the previous point], you can get the distance traveled by the rock, in the time you got here [part b of step 3], with the acceleration of gravity.

The key is to divide the problem in a horizontal projectile movement, and then a free-fall movement, using the horizontal distance as a total time measure.
Make a drawing. It should help.

Answer 3

The horizontal component h, of the initial velocity is constant.

h = 12.4*cos(60°) = 12.4*0.5 = 6.2 m/sec
t = time to travel horizontally 17 m.
t = distance/rate = 17 / 6.2 = 2.7419355 sec

v0 = initial vertical velocity

v0 = 12.4*sin(60°) = 12.4*(√3/2) = 10.738715 m/sec

v(t) = - 9.8t + v0

x = vertical position of rock

x(t) = ∫v(t) dt = ∫(-9.8t + v0)dt = -4.9t² + v0t

x(2.7419355) = -4.9(2.7419355)² + (10.738715)(2.7419355)
= -7.3943665 m

The height of the building is 7.3943665 m.