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

Answer:

$s_1 = 21.9 m$

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

$s = u t+ \dfrac{1}{2}at^2$

$s_1 = u t- \dfrac{1}{2}gt^2$

$s_2 = u (t - 1.5 )- \dfrac{1}{2}g(t - 1.5)^2$

for collision s₁ = s₂

$u t- \dfrac{1}{2}gt^2 = u (t - 1.5 )- \dfrac{1}{2}g(t - 1.5)^2$

$u t- \dfrac{1}{2}gt^2 = ut - 1.5 u - 0.5 g(t^2+ 2.25 - 3t)$

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

0 = -1.5 u + 1.5 gt -1.125 g

$t = \dfrac{1.5 \times 22 +1.125 \times 9.8}{1.5 \times 9.8}$

t = 2.99 s = 3 s

$s_1 = u t- \dfrac{1}{2}gt^2$

$s_1 = 22 \times 3 - 0.5 \times 9.8 \times 3^2$

$s_1 = 21.9 m$