A rock is thrown from a 20.0 m high cliff at an angle of 18.0° from the horizontal at a speed of 12.0 m/s. How long is the rock in the air before it hits the ground?

0.7560 seconds

Explanation:

The height of the cliff is 20 m.

The angle of projectile = 18°

The initial horizontal speed = 12 m/s

The formula to apply is :

T= {2*v*sinθ} /g  where;

T= time of flight of an object given initial launch angle θ and initial velocity v.

g= acceleration due to gravity

T= {2*12*sin 18°} / 9.81

T= {24 * 0.30901699437}/9.81

T= 0.7560 seconds

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