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 37º above the horizontal. Determine the the magnitude and the direction of the velocity at the maximum height.

Answer 1

Answer:

51.9 m/s, horizontal

Explanation:

The horizontal and vertical components of the initial velocity of the projectile are given by:

$v_x = v cos \theta=(65.0 m/s)(cos 37^{\circ})=51.9 m/s\\v_y = v sin \theta=(65.0 m/s)(sin 37^{\circ})=39.1 m/s$

In a projectile motion, the horizontal component of the velocity does not change (because there are no forces acting in the horizontal direction), while the vertical velocity changes according to:

$v_y(t)=v_{0y}-gt$

where g=9.8 m/s^2 is the acceleration due to gravity. However, at the point of maximum height, the vertical velocity is zero (because it is the point at which the projectile starts falling down), so the magnitude of the resultant velocity will simply be equal to the horizontal component:

v = 51.9 m/s

and the direction will be horizontal.