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a rocket is shot straight into the air an initial velocity of 96ft/sec. the height of the rocket after c sec. is represented by the formula

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4 Answers

Here is a bit of minutia that is more in the realm of physics.  If this was to be an absolutely precise measurement, you'd have to factor in the diminishing weight of the rocket due to the consumption of fuel.  Given a constant level of propulsion, a rocket will travel faster as its weight diminishes.

Hi, Maira.

The general equation h = 96x - 16x2 tells us the height at x seconds. We need to have a specific number of seconds to find the exact height.

For 1 second, plug 1 in for x:

h = 96(1) - 16(1)2    so     h = 80 ft

For 2 seconds, h = 96(2) - 16(4)  = 128 ft

3 seconds, h = 96(3) - 16(9)  = 144 ft

4 seconds, h = 96(4) - 16(16) = 128 ft

5 seconds, h = 96(5) - 16(25)  = 80 ft

6 seconds, h = 96(6) - 16(36)  = 0 , which means it has fallen back to the ground.

Hope the information here helps!

Maira, you did not copy the problem completely! I assume (from my experience) that you have to find the maximum height the rocket reaches.
Maximum height occurs when x = (- b) / (2a)
In our case, a = - 16, b = 96
    x = - 96 / [2 • (- 16)] = 3 seconds
  h(3) = 96 • 3 – 16 • 32 = 144 feet
The answer is (c)

I assume that the question is how high does it go, and that it decelerates at 32 ft/sec^2 (gravity) from time t=0. Then the highest point the rocket will reach is that point where its upward velocity reaches zero. Thereafter, the velocity goes negative (down) and the rocket loses height accordingly. Thus, since the acceleration reduces the upward velocity by 32 ft/sec every second, the time x when it becomes zero is t = 96/32 = 3 seconds. Plugging that into the formula for the height gives 96*3 - 16*3^2 = 96*3 - 16*9 = 288 - 144 = 144 ft. The answer is C.