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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 :

Answer:

[tex]s_1 = 21.9 m[/tex]

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

[tex]s = u t+ \dfrac{1}{2}at^2[/tex]

[tex]s_1 = u t- \dfrac{1}{2}gt^2[/tex]

[tex]s_2 = u (t - 1.5 )- \dfrac{1}{2}g(t - 1.5)^2[/tex]

for collision s₁ = s₂

[tex]u t- \dfrac{1}{2}gt^2 = u (t - 1.5 )- \dfrac{1}{2}g(t - 1.5)^2[/tex]

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

[tex]-0.5 gt^2 = -1.5 u - 0.5 gt^2+1.5 gt - 1.125 g[/tex]

0 = -1.5 u + 1.5 gt -1.125 g

[tex]t = \dfrac{1.5 \times 22 +1.125 \times 9.8}{1.5 \times 9.8}[/tex]

t = 2.99 s = 3 s

[tex]s_1 = u t- \dfrac{1}{2}gt^2[/tex]

[tex]s_1 = 22 \times 3 - 0.5 \times 9.8 \times 3^2[/tex]

[tex]s_1 = 21.9 m[/tex]

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