A ball is thrown vertically from the top of a bridge with an initial speed of 20m/sec.?

Question

On its way down, it just misses the top of the bridge and continues to fall to the ground.

a) How much time does it take the ball to reach the highest point?

b) How high is the highest point? What is the velocity of the ball at the end of 3 sec? If it takes 8 sec to reach the ground, how high is the bridge?

please try to solve...Thank you!

Answer 1

This is a one-dimentional problem so it's pretty straight forward. Sign convention is important, so assume "UP" is "+" and "DOWN" is "-" And I assume the 20m/s initial velocity is up....

a) At the highest point, the velocity will momentarily be zero, so:

Vf = 0 = Vi - g * t; solve for t = Vi / g = 20 / 9.8 = 2.04 sec

b) To find the height reached (relative to the bridge deck):

X = 1/2 * g * t^2 + Vi * t + Xo

X = 1/2 * (-9.8) * (2.04)^2 + 20 * 2.04 + 0; solve for

X = 20.4 m

c) The velocity after 3 sec is just:

V = Vi - g * t^2 = 20 - 9.8 * 3 = -9.4 m/s (it's headed back down)

d) To find the height of the bridge if total time is 8 sec, assume the bridge deck is Xo = 0m, and use

X = 1/2 * g * t^2 + Vi * t + Xo

X = 1/2 * (-9.8) * (8)^2 + 20 * 8

X = -153.6 m (153 m below the bridge deck reference)

Good Luck!

Answer 2

and has he throws the ball, the cute girl flirt at him.  He ignores the girls as he watches the ball.  He catches the ball and goes home without getting laid.