Question

A golf ball is hit from the ground with an initial speed of 192 feet per second. Assume the starting height of the ball is 0 feet. How long will it take the golf ball to hit the ground?

39 secs

19 secs

12 secs

6 secs

Answer 1

assume g = -32 , u = 192, s = 0, t = ?

using the formula

s = ut + 0.5gt^2
0 = 192t - 16 t^2
16t( t - 12) = 0
t = 12 seconds

Answer 2

Tysm

Answer 3

If you can help me with the rest of these Id be so grateful

Answer 4

i haven't a lot of time but give me one more

Answer 5

A juggler is performing an act by juggling several balls. The juggler throws the balls up at an initial height of 3.9 feet with a speed of 14.8 feet per second. If the juggler did not catch a ball, about how long will it take the ball to hit the floor?

1.45 secs

1.14 secs

0.28 secs

0.21 secs

Answer 6

again g = -32 ft per sec, s = displacement = -3.9, u = 14.8, t = ?

-3.9 = 14.8t - .5 * 32 * t^2
16t^2 - 14.8t - 3.9 = 0
t = -o.21 or 1.14
time cant be negative so t must be 1.14 secs

Answer 7

Thats what I thought it was cool

Answer 8

yeah you need the constant acceleration equations to solve these type problems

they are

v = u + at v = final velocity, u = initial , a = acceleration ( g= accn of gravity)
s = ut + 0.5at^2
v^2 = u^2 + 2as
s =[ (u+v)t] / 2

s = displacement