the height in meters of an object above the ground at time t

the height in meters of an object above the ground at time t is given by the equation h = 15t - 4.9 t^2, where 15 m/s is the initial velocity.

a) at what height is the object initially?

b) how long is the object in the air before it hits the ground?

c) when will the object reach its maximum height and how high above the ground will it be?

Solution

h = 15*t - 4.9t^2

a)

At t = 0

h = 0 <---- It is at 0 m initially

b)

h = 15*t - 4.9t^2 = 0

So, t = 15/4.9 = 3.06 s <----answer

c)

For Max height = dh/dt = 0

So, 15 - 4.9*2*t = 0

So, t = 1.53s <-----answer time for max height

And, maximum height:

H = 15*1.53 - 4.9*1.53^2 = 11.5 m <---answer