- #1
VaioZ
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Homework Statement
A ball is thrown vertically upward with an initial speed of 19.6 m/s from the top of a building 44.1 m tall. Assuming that there is no air resistance.
a.) At what two times is the ball 10.0 m above its starting point?
b.) What is the position of the ball at t = 5.00 s?
c.) How long will it take the ball to strike the ground?
d.) What is the velocity of the ball when it strikes the ground?
Homework Equations
Kinematics Equation/Projectile Equations
The Attempt at a Solution
a.) First I took 10 m as my delta Y then 19.6 m/s as may initial velocity to get the time
10 m = (19.6 m/s)(t) - (1/2)(9.8m/s^2)(t)^2
t = 3.3997 s
b.) First I took the time to get to the max height which is (19.6 m/s)/(9.8 m/s^2) = 2 s as my Tmax
then I compute for the Ymax which is Y=(19.6 m/s)(2 s) - (1/2)(9.8 m/s^2)(2 s)^2
Ymax = 19.6 m
then I added 19.6 m + 44.1 m = 63.7 m to get the height from base of the building to the max height of the projectile
then next is to get the position of the ball at 5 s
Y=(19.6 m/s)(5 s) - (1/2)(9.8 m/s^2)(2 s)^2
Y=-24.5 m
Then I substract 63.7 m - 24.5 m = 39.2 m
c.) 63.7 m = (19.6 m/s)(t) - (1/2)(9.8)(t)^2
but then it can't be solved sooooo I think I got wrong somewhere
pls help thank you.