Break the velocty into vertical and horizontal components.
Then, height=1.2 + vivertical*t-4.9t^2
and hmax occurs at the t when
Vv=0=vivertical -9.8t
A ball is thrown from a point 1.2 m above the ground. The initial velocity is 19.8 m/s at an angle of 34.0° 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.
1 answer
1 answer