Question

A heavy rock is shot upward from the edge of a vertical cliff. It leaves the edge of the cliff with an initial velocity of 12 m/s directed at 26° from the vertical and experiences no appreciable air resistance as it travels. The distance the rock landed was 40.4 m away from the cliff. How high is the cliff?

Answer 1

Answer:

The height of the building is 88.63 m.

Explanation:

Given;

initial component of vertical velocity, $v_i$ = 12 m/s sin 26° = 5.26 m/s

initial horizontal component of the velocity, $u_i$ = 12 m/s cos 26° =10.786 m/s

horizontal distance traveled by the rock, x = 40.4 m

time of flight is calculated as;

x = $u_i$ t

t = x / $u_i$

t = 40.4 / 10.786

t = 3.75 s

Determine the final vertical velocity of the ball;

$v_f = v_i + gt\\\\v_f = 5.26 + (9.8 *3.75)\\\\v_f = 42.01 \ m/s$

Determine the height of the rock;

$v_f^2 = v_i^2 + 2gh\\\\h = \frac{v_f^2 - v_i^2}{2g}\\\\ h = \frac{(42.01)^2 - (5.26)^2}{2*9.8}\\\\h = 88.63 \ m$

Therefore, the height of the building is 88.63 m.