Question

Two toy cannons are mounted right next to each other. The first cannon launches a small ball upward, with an initial velocity of 22.0 m/s. 1.5 seconds later, the second cannon launches a second ball upward, also at 22.0 m/s. The two balls collide in the air. Find the height above the cannons where the balls collide.

Answer 1

`s_1 = 21.9 m`

Step-by-step explanation:

given,

initial velocity of the first cannon = 22.0 m/s

after 1.5 s second cannon launches at speed = 22 m/s

height where they collide = ?

using equation of motion

`s = u t+ (1)/(2)at^2`

`s_1 = u t- (1)/(2)gt^2`

`s_2 = u (t - 1.5 )- (1)/(2)g(t - 1.5)^2`

for collision s₁ = s₂

`u t- (1)/(2)gt^2 = u (t - 1.5 )- (1)/(2)g(t - 1.5)^2`

`u t- (1)/(2)gt^2 = ut - 1.5 u - 0.5 g(t^2+ 2.25 - 3t)`

`-0.5 gt^2 = -1.5 u - 0.5 gt^2+1.5 gt - 1.125 g`

0 = -1.5 u + 1.5 gt -1.125 g

`t = (1.5 * 22 +1.125 * 9.8)/(1.5 * 9.8)`

t = 2.99 s = 3 s

`s_1 = u t- (1)/(2)gt^2`

`s_1 = 22 * 3 - 0.5 * 9.8 * 3^2`

`s_1 = 21.9 m`