Question

4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strikes the ground at a horizontal distance of 160 m from the base of the building 5.0 s after being thrown. Determine the speed with which the rock was thrown.

Answer 1

Answer:

Explanation:

Let the velocity of projectile be v and angle of throw be θ.

The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

considering its vertical displacement

h = - ut +1/2 g t²

100 = - vsinθ x 5 + .5 x 9.8 x 5²

5vsinθ =  222.5

vsinθ = 44.5

It covers 160 horizontally in 5 s

vcosθ x 5 = 160

v cosθ = 32

squaring and adding

v²sin²θ +v² cos²θ = 44.4² + 32²

v² = 1971.36 + 1024

v = 54.73 m /s

Answer 2

Answer:

55.42 m/s

Explanation:

Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by

u_x = d/t

Where

d = 160 m is the distance covered

t = 5.0 s is the time taken

Substituting, we get

u_x =160/5 = 32 m/s.

Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the

$S=u_yt+\frac{1}{2}at^2$

where

S = -100 m is the vertical displacement

u_y is the initial vertical velocity

Replacing t = 5.0 s and solving the equation for u_y, we find

-100 = u_y(5) + (-9.81)(5)^2/2

u_y = 45.25 m/s

Therefore, the speed with which the rock was thrown u

$u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s$