Question

A 2kg ball is launched from a height of 1.5m, with an initial velocity of 8.0ms^-1. The launch angle is 60° to the horizontal. What is the maximum height from the ground that is reached by the ball?

Answer 1

Mass ball:

m=2kg

Initial height:

h_0=1.5m

Initial velocity:

V=8m/s

Angle:

α=60°

Gravity:

g=9.8 m/s^2

Solution:

The weight of the ball in the calculation is not involved.

According to the law of motion:

h=h_0+Vt*sinα-(gt^2)/2

The height will be greatest if the vertical speed is 0:

0=Vsinα-gt

We get the time to reach the maximum height:

t=Vsinα/g

From here we get the formula of the greatest height:

h=h_0+V^2*(sinα^2)/2g=1.5+8^2*〖0.866〗^2/(2*9.8)=1.5+2.45=3.95m