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[tex]\theta[/tex]/g
U = velocity of the ball
[tex]\theta[/tex] = 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 [tex]\theta[/tex] will be 0° in this case
If Ux = Ucos[tex]\theta[/tex]
8 = Ucos0
U = 8m/s
Since t = Usin[tex]\theta[/tex]/g
t = 8sin0/9.8
t = 0sec
