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

Physics

1 answer:

vovangra [49]3 years ago

8 0

Answer:

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

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