A toy rocket is launched from the top of a building 95 feet tall at an initial velocity of 201 feet per second.

Question

​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

513

feet above ground​ level?

​d) After how many seconds will it hit the​ ground?

Answer 1

95+201t=h

Need more imfo power/time.

513-95=418/201=2.08seconds

gravity/down=(t^2+t)/2*32.2=If 513' is the max=5.167sec.

Answer 2

(a) Well we use the motion equation from physics s = ut + 0.5 at²

^{s = distance above top of building (ft) , u = initial vertical speed (ft/s), a = acceleration due to gravity ( - 32 ft/s/s), negative as it is downwards}

So when t = 0, s = 0

u = 201

a = - 32

So height of rocket above top of building will be 201 t - 0.5 x 9.81 x t² = 201t - 0.4905 t², after t seconds

Height above ground will be 95 + 201 t - 0.4905 t²

^(b). Draw the function in (a) on Desmos or a graphing calulator and read off the time (x axis) and height (y axis) at the highest point (vertex)

(c) Read this off your graph

(d) Read this off your graph

For (c) you can also solve 513 = 95 + 201t - 0.4905t²

For (d) you can solve 95 + 201t - 0.4905t² = 0

Mike