Question

The height in meters of a projectile launched from the top of a building is given by h(t)=-5t^2+60t+15, where t is time in seconds since it was launched.
a) How tall is the building?
b) At what time does the projectile hit the ground?
C)What is the velocity(rate of change with respect to time) of the object when it hits the ground?
d)Determine the projectile's velocity at 2 seconds.

Answer 1

YOU have gravity reversed !!!!

the building is 15 meters high since that is h when t = 0

Answer 2

h = -5 t^2 + 60 t + 15

at ground
0 = -5 t^2 + 60 t + 15

t^2 - 12 t - 3 = 0

t^2 -12 t = 3

t^2 - 12 t + 36 = 39

(t-6)^2 = 39

t - 6 = +/- sqrt 39
t = 6 + 6.24 = 12.24 seconds

dh/dt = -10 t + 60
= -10(12.24) + 60
= -62.4 m/s

dh/dt =-10 t + 60 at 2 seconds
= -20 + 60
= 40 m/s

Answer 3

Thank you Damon :)