Question

A projectile leaves the muzzle of a projectile launcher with a speed of 35m/s [the

launcher is oriented vertically].

a) What is the maximum height reached by the projectile?

b) How long did it take the projectile to reach its maximum height?

c) How long was the projectile in the air [ttot]?

Answer 1

a. 62.44 m

b. 3.57 sec

c. 7.12 sec

Step-by-step explanation:

The leaving speed is given as = 35 m/s

The launcher is oriented vertically, thus the angle is = 0 degrees

a) The maximum height reached is given as;

h= v²₀y / 2g

h=35² / 2 * 9.81

h= 62.44 m

b) Time the projectile reaches maximum height is given as;

t=v₀y / g

t=35/9.81

t= 3.57 sec

c) The trip up is a mirror of the trip down

This means the time the projectile will be in the air is ;

t= 2 * 3.57

t= 7.12 sec