Question

A projectile is fired at an upward angle of 35.6° from the top of a 208-m-high cliff with a speed of 170-m/s. What will be its speed (in m/s) when it strikes the ground below?

Answer 1

Answer:

Final velocity is 181.61 m/s at angle 40.44° below horizontal.

Explanation:

Initial horizontal velocity = 170 cos 35.6 = 138.23 m/s

Final horizontal velocity = 138.23 m/s

Considering vertical motion of projectile:

Initial vertical velocity, u = 170 sin 35.6 = 98.96 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = -208 m

We have v² = u² + 2as

Substituting

             v² = 98.96² + 2 x -9.81 x -208

             v = 117.79 m/s

Final velocity,

            $v=\sqrt{138.23^2+117.79^2}=181.61m/s$

            $\theta =tan^{-1}\left ( \frac{117.79}{138.23}\right )=40.44^0$

Final velocity is 181.61 m/s at angle 40.44° below horizontal.