Question

A 0.90-kg ball is thrown with a speed of 9.0 m/s at an upward angle of 27 ∘.A) what is its speed at its highest point, andb)how high does it go?

Answer 1

Answer:a)8.01 m/s

Step-by-step explanation:

Given

mass of ball
`m=0.9 kg`

initial speed
`u=9 m/s`

launch angle
`\theta =27`

As the projectile reaches its highest point its vertical velocity component becomes zero and there will only be horizontal component

velocity at highest Point
`v=u\cos \theta`

`v=9\cos 27`

`v=8.01 m/s`

(b)maximum height h

Maximum height is given by

`H_(max)=(u^2\sin^2 \theta )/(2g)`

`H_(max)=(9^2\sin^2 (27))/(2* 9.8)`

`H_(max)=0.85 m`