Question

A rocket is launched straight up from the side of the cliff 43.89m high if the initial velocity is 34.14 m/sec qnd the height is givenby the formula H=-16
tsquare + vt + h. Find time at which the rocket will hit the ground

Answer 1

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Answer 2

Given: A rocket is launched straight up from the side of the cliff 43.89m high if the initial velocity is 34.14 m/sec qnd the height is givenby the formula H=-16 tsquare + vt + h.

To find: The time at which the rocket will hit the ground.

Solution:

According to the question, the height of the rocket is given by the following formula.

$H = -16t^{2} + vt + h$

Here, H is the final height of the rocket, t is the time taken, v is the initial velocity and h is the initial height of the rocket. When the rocket hits the ground its height will be 0 m. Now, the formula can be represented as follows.

$0 m = -16t^{2} + (34.14 ms^{-1})t + 43.89$

$16t^{2} - 34.14 t - 43.89 = 0$

On solving the quadratic equation, the values of t are found to be

$t = 3.04 s$ or $t = - 0.90 s$

Since the time taken cannot be negative, the time taken for the rocket to hit the ground is 3.04 s.

Therefore, the time at which the rocket will hit the ground is after 3.04 s.