A symmetric projectile spends a total of 6 seconds in the air. How long does it take to reach its peak height? The projectile has a starting velocity of 8m/s. What is its horizontal range? What is the projectile's horizontal acceleration? What is the projectile's vertical acceleration?

Given a Projectile motion

The total time of flight T=6s

a. Time to reach maximum height=?

The time to reach maximum height is half of the time of flight given,

Then,

t=T/2

t=6/2

t=3s

The time to reach maximum height is 3sec.

b. If the horizontal initial velocity is 8m/s

Ux=8m/s

The range of a projectile is given as.

R=Ux•T

Given that T=6s and Ux, = 8m/s

R=Ux•T

R=8*6

R=48m.

The range of the projectile is 48m.

c. The horizontal acceleration is zero.

This is due to no horizontal force acting on the projectile. So, there is no horizontal acceleration. That proposition of no horizontal acceleration is approximately correct in our world.

d. The vertical acceleration is g=9.81m/s²

The Gravity is the downward force upon a projectile that influences its vertical motion.

So the vertical acceleration is 9.81m/s²