Question

Using the projectiles equations for y and x at what second does the ball have a vertical velocity of -39.2 m/s^2

Answer 1

a) t = 4

Step-by-step explanation:

In projectile the time taken by an object to reach its maximum height is the maximum time (tmax).

tmax = Usin
`\theta`/g

U = velocity of the ball

`\theta` = angle of projection = 90° (vertical velocity)

g = acceleration due to gravity = 9.8m/s²

tmax = Usin90°/9.8

tmax = U/9.8

T the maximum height U = Uy = 39.2 m/s²

tmax = 39.2/9.8

tmax = 4secs

To calculate the time at which the ball will have an horizontal velocity, we will use the same formula but
`\theta` will be 0° in this case

If Ux = Ucos
`\theta`

8 = Ucos0

U = 8m/s

Since t = Usin
`\theta`/g

t = 8sin0/9.8

t = 0sec