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Two rockets are launched at a fireworks display. Rocket A is launched with an initial velocity v0 5 100 m/s and rocket B is launched t1 seconds later with the same initial velocity. The two rockets are timed to explode simultaneously at a height of 300 m as A is falling and B is rising. Assuming a constant acceleration g 5 9.81 m/s2, determine (a) the time t_1, (b) the velocity of B relative to A at the time of the explosion.

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Answer

given,

initial speed of the rocket A = 100 m/s

height of explode = 300 m

acceleration due to gravity = 9.8 m/s²

rocket b is launched after t₁ time

now, using equation of motion to calculate time

[tex]s = ut + \dfrac{1}{2}gt^2[/tex]

[tex]300 = 100t + \dfrac{1}{2}(-9.8)t^2[/tex]

 4.9 t² - 100 t + 300 = 0

using quadratic equation

[tex]t = \dfrac{-(-100)\pm \sqrt{100^2-4\times 4.9 \times 300}}{2\times 4.9}[/tex]

t₁ = 3.65 s   and  t₂ = 16.75 s

now, rocket A reaches 300 m on return at 16.75 s

rocket B reaches 300 m after 3.65 s

time difference of launch:

t = 16.75 - 3.65

t = 13.1 s

velocity of rocket A

v_a = u + g t

v_a =100 - 9.8 x 16.75

v_a = -64.15 m/s

velocity of rocket B

v_b = u + g t

v_b =100 - 9.8 x 3.65

v_b =+64.23 m/s

relative velocity of B relative to A at the time of the explosion

[tex]V_{BA} = v_b - V_a[/tex]

[tex]V_{BA} = 64.23 -(-64.15)[/tex]

[tex]V_{BA} = 128.38\ m/s[/tex]

relative velocity is equal to 128.38 m/s