Question

A ball is kicked straight up from ground level at a velocity of 13 m/s. Find:
a) the maximum height
b) the time at which the ball reaches maximum height
c) the time required for the entire trip
d) the ball's velocity just before it hits the ground upon returning

Answer 1

a) The kinetic energy given to the ball is 1/2MVo^2 and when the ball has reached the maximum height, all this energy has turned into potential energy Mgh, then solve: 1/2MVo^2=Mgh or 1/2Vo^2=gh---->>>>h=Vo^2/(2g)
2)Apply V=Vo-gt and calculate t when V=0 (maximum height) t=Vo/g
3)h=1/2gt^2 and h=Vo^2/(2g) ----->>>t=Vo/g, then t(total)=2Vo/g
4)When the ball hits the ground, all its potential energy has turned into kinetic energy. Assuming that no energy has been lost in the trip (due to friction with air, for instance), the speed has to be the same with which we kicked it up

Answer 2

I really don't get this because I haven't began to use the equations that you have provided. I'm currently using equations involving uniform motion with initial velocity, final velocity, distance, time and acceleration.

Answer 3

1. solve for time (t) when velocity = 0. ( 0 = -9.81*t +13) ----> t = 1.325 seconds


2. solve for position when t = 1.325 seconds y = 0.5*(-9.81)(1.325)^2 + 13*1.325 ----> y = 8.61 m

3. without friction, the time for the entire trip is 2* the time for half the trip (when the ball is at its peak height). so the time of the entire trip is 2.65 seconds.


4. without friction, the velocity just before the ball hits the ground is the same as its initial velocity, 13 m/s.

Answer 4

@Daenio
The equations you mention are used in 2) & 3). For 1) and 4), here it goes
For 1) At h max, v=0 and v=Vo-g*t---->>>t=Vo/g. Now replace this time in the equation: h(max)=Vo*t-1/2*g*t^2=Vo*(Vo/g)-1/2*(Vo/g)^2*g=Vo^2/(2g) (same result than when using energy)
4)Falling from h max is h=h(max)-1/2*g*t^2--->If h=0 (back to ground) then t^2=2*h(max)/g---->t^2=2*[Vo^2/(2g)]/g=Vo^2/g^2----->t=Vo/g and V=g*t=Vo/g*g=Vo (same result than when using energy)

Answer 5

\[V_0=13m/s \space \space; \space \space g =10m/s^2\]
I part .) \[h_{\max.}=\frac{V_0^2}{2g}.\]
II part.)\[h_{\max.}=V_0t+\frac{1}{2}g t^2\]So find "t"

III part.)
Time required for entire trip = 2 x t.

IV part.)
the ball's velocity just before it hits the ground upon returning = Vo = 13m/s

Answer 6

Thank you!