Problem 61.

A ball is thrown from a point 1 m above the ground.  The initial velocity is 20 m/s at an angle of 40 degrees above the horizontal.

(a) Find the maximum height of the ball above the ground.

(b) Calculate the speed of the ball at the highest point in the trajectory.

Solution:

This is the projectile motion. The equation which describe the motion along horizontal axis x is the following

where is the x-component of initial velocity.

Along axis y we have motion with constant acceleration:

(a) The point of the maximum height corresponds to the condition that . Then

and

Then

This is the maximum height.

(b) The y-component of velocity at maximum height is 0, then the speed is equal to x-component of velocity: