A projectile is shot from the edge of a cliff 125 m above ground level with an i

Question

A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37.0 with the horizontal, as shown below. (a) Determine the time taken by the projectile to hit point P at ground level. (b) Determine the range X of the projectile as measured from the base of the cliff. At the instant just before the projectile hits point P, find (c) the horizontal and the vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal. (f) Find the maximum height above the cliff top reached by the projectile. please show the equations and steps

Explanation / Answer

At time t after launch, the projectile's co-ordinates measured from ground level are: x = 65.0t cos(37.0) ...(1) y = 125 + 65.0t sin(37.0) - 9.81t^2 / 2 ...(2) Putting y = 0 and solving (2) for t: 9.81t^2 / 2 - 65.0 t sin(37.0) - 125 = 0 t = (65.0 sin(37.0) +/- sqrt(65.0^2 sin^2(37.0) + 2 * 9.81 * 125)) / 9.81 = 10.4 sec. (b) Substituting this value of t in (1): x = 541 m. (c) x' = 65.0 cos(37) = 51.9 m/s. y' = 65.0 sin(37) - 9.81 * 10.4 = - 63.1 m/s. upward = 63.1 m/s downward. (d) sqrt(51.9^2 + 63.1^2) = 81.6 m/s. (e) If a is the angle, tan(a) = y' / x' a = 50.6 deg. (f) If h is the height: 0 = 65.0^2 sin^2(37.0) - 2 * 9.81h h = 78.0 m.

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