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Musya8 [376]
2 years ago
11

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?
Physics
1 answer:
jenyasd209 [6]2 years ago
5 0

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.

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Answer:

p = 6.64 cm

Explanation:

For this exercise we use the equation of the constructor

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