Can someone help me with these physics problems on freefall?

Question

Can you also show work and explain how you got the answer thanks!

I just don't get how you are suppose to solve the problem with 2 unknowns


A foul ball is hit straight up into the air with a speed of about 22 m/s.
(a) How high does it go?
m
(b) How long is it in the air?
s

no, 9.8 m/s is the acceleration

Answer 1

you should use these two equations to solve the problem

x' = (v_f^2-v_o^2)/2a
v_f = at + v_o

where v_f equals final velocity = 0, and v_o = 22 m/s

x' = displacement traveled is what we need to find for part (a)

x' = (0-22^2)/(2*9.8) = 24.7 meters (ignore negative sign)

part (b), use 2nd equation

0 = 9.8t + 22

t = 22/9.8 = 2.24 seconds (ignore negative sign).

It take 2.24 seconds to reach the peak of 24.7 meters, but it takes another 2.24 seconds to reach the ground.

Thus, total time in air = 4.48 seconds

Answer 2

Here are the equations: a = consistent = -nine.eight m/s² (continually) V = Vo + a*t Y = Yo + Vo*t + ½*a*t² (known as 'function' within the query) Vo is +nine.eight m/s and Yo is 0 on the best of the constructing. Put the time values into the above equations and fill within the chart. The solutions to the two ?s under it'll turn out to be obvious. The best of the flight is whilst V = zero and the roof stage implies Y = zero

Answer 3

Speed=acelleration*time, time(s)=22/.9.8=2.24 sec (*2, up and down, =4.48sec)
Distance=speed*time, =1/2*2.24*22=24.7m
(with constant ac/de-celleration, average speed is 1/2 of top speed)

Of course real numbers will be different due to air resistance/changes in gravity. (higher you go less it gets)

Answer 4

The second unknown is probably gravity (a) , which equals 9.81 m/s^2.