Height of the ball above ground level

Question

John throws a ball straight up into the air at 50 ft/sec while standing on top of a 300ft tower.

a) write an algebraic expression that describes the height of the ball above ground level t seconds after it is thrown

b) how far above the ground is the ball 4 seconds after it is thrown

c) when does the ball hit the ground? round your answer to the nearest hundredths.

Answer 1

H = h0 +Vi*t -1/2 gt²

H= 300 +50t - 1/2 gt²

Height after 4 sec

H = 300 +50*4 -1/2 ( 32) (4*4) = 244 ft (using g = 32 ft/s2 )

Time it hits the ground

H= 0

0 = 300 +50t -1/2 gt²

300 +50t - 16t² = 0

16t2 -50t -300 = 0

t = 6.16 s