The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of 24.5 m/s is h = 4 + 24.5t − 4.9t2 after t seconds.

Question

(a) Find the velocity after 2 s and after 4 s

d) When does it hit the ground?

(e) With what velocity does it hit the ground?

I just need help with d and e

Answer 1

It hits the ground when h = 0

0 = −4.9t² + 24.5t + 4

Using the quadratic equation with a = -4.9, b = 24.5, and c = 4

t = [-b ± √(b² - 4ac)]/(2a)

t = [-24.5 ± √((24.5)² - 4(-4.9)(4)]/(2(-4.9))

t = [-24.5 ± √(600.25+78.4)]/(-9.8)

t = [-24.5 ± √678.65]/(-9.8)

t = 5.16 sec or t = -0.16 sec

Obviously, we use the 5.16 sec answer

The velocity is dh/dt = 24.5 - 9.8t and at t = 5.16, it is -26.05 m/s (going downward at 26.05 m/s)