Question

A ball is thrown vertically upward from top of a building 96m tall with an initial velocity 80m / second. The distance 's' of the ball from the ground after t seconds is S = 96 + 80t - 4.9t2. After how many seconds does the ball strike the ground.

Answer 1

A ball is thrown vertically upward from the top of a building 96feet tall with an initial velocity of 80feet per second. The distance S (in feet )of the ball from the ground, after t seconds is s=96+80t-16tsquard.
(a)After how many seconds does the ball strike the ground?
When s = 0 it hits the ground
s=96+80t-16t^2 = 0
-t^2 + 5t + 6 = 0
(-t + 6)*(t + 1) = 0
t = -1 second (Ignore)

t = 6 seconds to impact
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(b) After how many seconds will the ball pass the top of the building on its way down?
96+80t-16t^2 = 96
-t^2 + 5t = 0
t = 0 (at launch)
t = 5 seconds (on the way down)

Answer 2

Answer:

Solve the equation

Step-by-step explanation:

96+80t-16t^2

Answer is 6s