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

Answer:

a) t = 4

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