Question

a ball is thrown upward from the top of a 98 m. tower with initial speed 39.2 m/s. how much later will it hit the ground ?

Answer 1

hint : consider the top of the tower as level zero . if h is the height of the ball above the top of the tower , then h=-98 when the ball hits the ground/

Answer 2

Use the distance-velocity formula

Answer 3

acceleration from gravity = 9.8 m/s^2, I think.

Answer 4

h=rt-4.9t ^2

Answer 5

So what is h going to be?

Answer 6

Should be -98

Answer 7

Sorry. Right.

Answer 8

98 = 39.2t - 4.9t^2

becomes

4.9t^2 - 39.2t + 98 = 0

Now solve for t.

Answer 9

Plug into the quadratic formula..

Answer 10

Yes, -98

Answer 11

@crombie use his formula, except correct it to -98, OK?