A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from theground after t seconds is s(t)= 96t - 16t^2 A) At what time t will the ball strike the ground? B) For what time t is the ball more than 128 feet above theground?

grippeb9

grippeb9

Answered question

2022-08-07

A ball is thrown vertically upward with an initial velocity of96 feet per second. The distance s (in feet) of the ball from theground after t seconds is s ( t ) = 96 t 16 t 2
A) At what time t will the ball strike the ground?
B) For what time t is the ball more than 128 feet above theground?

Answer & Explanation

Winston Cooper

Winston Cooper

Beginner2022-08-08Added 12 answers

A) the distance of the ball from the ground after t sec is given by the equation,
s ( t ) = 96 t 16 t 2
at the time of striking the ground the distance of the ball from ground = 0 feet
so, 96 t 16 t 2 = 0
16 t 2 96 = 0
16 t ( t 6 ) = 0
hence either, 16t =0 or t - 6 = 0
t =0 or t = 6
t=0 when the ball is thrown
t = 6 when the ball reaches the ground afterthrowing.
Answer: after 6 seconds of throwing the ball it will reach the ground.
B) The distance of the ball from the ground after t sec is given by the equation,
s ( t ) = 96 t 16 t 2
at the time when the ball is 128 feet above the ground the distanceof the ball from ground = 128 feet.
so, 96 t 16 t 2 = 128
16 t 2 96 t + 128 = 0
16 ( t 2 6 t + 8 ) = 0
t 2 6 t + 8 = 0
t 2 4 t 2 t + 8 = 0
t ( t 4 ) 2 ( t 4 ) = 0
( t 4 ) ( t 2 ) = 0
hence either, t - 4 =0 or t - 2 = 0
t = 4 or t = 2
t = 4 when the ball is moving downwards to ground.
t = 2 when the ball is moving to upward direction afterthrowing.
Answer: after 2 and 4 seconds of throwing the ball it will reach the ground.

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