A Projectile. When a projectile is launched upward from the top of a building of height H, its vertical position (in feet) at time t (in seconds) is given by

Question

A Projectile. When a projectile is launched upward from the top of a building of height H, its vertical position (in feet) at time t (in seconds) is given by. Show Work

s = -16t^2 + 96t +H

a) Find the initial velocity:

b)  How long does it take for the projectile to reach its maximum height?

c) How long does it take for the projectile to return to its original height?

D) The projectile hits the ground 10 seconds after being launched. What is its final velocity when it hits the ground?

e) Find the height of the building, H.

Answer 1

Hi,

s = -16t^2 + 96t +H =====> a = -16 b = 96 c =H

a) Find the initial velocity:

velocity = ds/dt = -32t + 96

b)  How long does it take for the projectile to reach its maximum height?

t = -b/2a ====> t = -96/2(-16) = 96/32 = 3 after 3 second it reaches to the maximum height

c) How long does it take for the projectile to return to its original height?

s = -16t^2 + 96t +H if we assume the original height is H as a result we have s = -16t^2 + 96t +H = H

-16t^2 + 96t +H = H -----> -16t^2 + 96t = 0 -----> -16t( t - 6) = 0 t = 0 , 6 so after 6 second it returns to the original height

D) The projectile hits the ground 10 seconds after being launched. What is its final velocity when it hits the ground?

V(10) = -32 ( 10) + 96= -320 + 96

I hope it is useful,

Minoo