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
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 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
Beginner2022-08-08Added 12 answers
A) the distance of the ball from the ground after t sec is given by the equation,
at the time of striking the ground the distance of the ball from ground = 0 feet so,
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,
at the time when the ball is 128 feet above the ground the distanceof the ball from ground = 128 feet. so,
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.