a. hmax = ho + (Y^2-(Vo*sin33.8)^2)/2g.
hmax = 1.38 + (0-122.6)/19.6 = 6.26 m.
b. Velocity is zero at max. ht.
A ball is thrown from a point 1.38 m above the ground. The initial velocity is 19.9 m/s at an angle of 33.8° above the horizontal. Find the maximum height of the ball above the ground.
Calculate the speed of the ball at the highest point in the trajectory.
1 answer
1 answer