Question

A baseball is thrown with a velocity of 15 m/s at an angle of 30 degrees ccw from the horizontal.

a) What are the horizontal and vertical components of the velocity just as the ball leaves the hand?

b) What is the horizontal and vertical velocity of the ball when it reaches maximum height?

c) At what time does the ball reach maximum vertical height?

d) How far has the ball traveled vertically and horizontally at this time?

Answer 1

a) horizontal velocity = 15 cos 30 = 13 m/s

Vertical velocity = 15 sin 30 = 7.5 m/s

b) at maximum height ,

vertical velocity = 0

horizontal velocity = 13 m/s

c) time taken to reach the maximum height = 15 sin 30 / 9.8 = 0.76 s

d) vertical movement = u^2 sin^2 a / 2g = 15^2* (sin 30)^2 / 2*9.8 = 2.87 m

horizontal movement = R/2 = u^2 sin 2a / 2g = (15)^2 * sin 2*30 / 2*9.8

= 9.94 m