A projectile is shot from the edge of a cliff 205 m above ground level with an i

Question

A projectile is shot from the edge of a cliff 205 m above ground level with an initial speed of vo=135 m/s at an angle of 35.0° with the horizontal, as shown in the figure.

(a) Determine the time taken by the projectile to hit point P at ground level.
s

(b) Determine the distance X of of point P from the base of the vertical cliff.
km

(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Assume the positive directions are upward and to the right.)
m/s (horizontal)
m/s (vertical)

(d) At the instant just before the projectile hits point P, what is the magnitude of the velocity?
m/s

(e) At the instant just before the projectile hits point P, what is the angle made by the velocity vector with the horizontal?
° below the horizontal

(f) Find the maximum height above the cliff top reached by the projectile.
m

Explanation / Answer

(a) Determine the time taken by the projectile to hit point P at ground level. The velocity of the projectile along the vertical direction is Using basic relations we have v^2 = u^2 + 2as v^2 = (135 sin35)^2 + 2(-9.8 )(-205) |v| = 63.99 m/s but v must be going down so v = -63.99 m/s The time taken by the projectile is v = u + at ==> -63.99 = 135sin35 - 9.8 t ==> t = 14.43 s (b) Determine the distance X of of point P from the base of the vertical cliff. X = u cos@ * t = 135 cos35 * 14.43 = 1596 m (c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Assume the positive directions are upward and to the right.) m/s (horizontal) m/s (vertical) Vx = 135 cos35 = 110.58 m/s Vy = |v| = -63.99 m/s (d) At the instant just before the projectile hits point P, what is the magnitude of the velocity? |V| = 127.76 m/s (e) At the instant just before the projectile hits point P, what is the angle made by the velocity vector with the horizontal? ° below the horizontal @ = tan-1 (-63.99 / 110.58) = 30.05 deg

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