Question

A toy rocket is launched from the top of a building 94 feet tall at an initial velocity of 242 feet per second. ​a) Give the function that describes the height of the rocket in terms of time t. ​b) Determine the time at which the rocket reaches its maximum​ height, and the maximum height in feet. ​c) For what time interval will the rocket be more than 419 feet above ground​ level? ​d) After how many seconds will it hit the​ ground?

Answer 1

Assuming a parabolic movement we have to,

`h (t) = h + vt+ (at^2)/(2)`

Gravity is equivalent to 32.2ft / s ^ 2

So,

a)
`h (t) = 94 + 242t - 0.5 (32.2) t ^ 2`

b) h_max would be the vertex of the parabola

`t_(max) = -(b)/(2a) = -242 / (- 32.2) = 7.5155 sec`

**Here a=g/2

`h_(max) = h (7.5155) = 94 + 242 (7.515) - (0.5) (32.2) * (7.5) ^ 2 = 1003.38 ft`

c)
`h (t) = 419 = 94 + 242t - 0.5 (32.2) t ^ 2`

Solve for t using quadratic formula:

1.49 s <= t <= 13.54 s

d) Solve for t:
`h (t) = 94 + 242t - 0.5 (32.2) t ^ 2 = 0`

`t = 15.4099 s`