Question

Throw a small ball (best a golf ball) straight upward as hard as you can. Your friend can measure the time of flight of the ball using a stop watch.
Find the initial speed of your throw and the approximate maximum height of the ball solely with the use of the measured time.
you are given four times : (all in seconds) 2.20 2.43 2.74 2.14
what is the vo & hmax given these formulas for each time:

use Time of flight = 2

Answer 1

1) for T = 2.2 s

T = 2*Vo*sin(theta)/g

2.2 = 2*Vo*sin(90)/9.8

vo = 2.2*9.8/2

= 10.78 m/s

Hmax = vo^2*sin^2(theta)/(2*g)

= 10.78^2*sin^2(90)/(2*9.8)

= 5.93 m

2)for T = 2.43 s

T = 2*Vo*sin(theta)/g

2.43 = 2*Vo*sin(90)/9.8

vo = 2.43*9.8/2

= 11.907 m/s

Hmax = vo^2*sin^2(theta)/(2*g)

= 10.78^2*sin^2(90)/(2*9.8)

= 7.23 m

3)
for T = 2.74 s

T = 2*Vo*sin(theta)/g

2.74 = 2*Vo*sin(90)/9.8

vo = 2.74*9.8/2

= 13.426 m/s

Hmax = vo^2*sin^2(theta)/(2*g)

= 10.78^2*sin^2(90)/(2*9.8)

= 9.197 m

4)for T = 2.14 s

T = 2*Vo*sin(theta)/g

2.14 = 2*Vo*sin(90)/9.8

vo = 2.14*9.8/2

= 10.486 m/s

Hmax = vo^2*sin^2(theta)/(2*g)

= 10.78^2*sin^2(90)/(2*9.8)

= 5.61 m

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2) we have, T = 2vo*sin(theta)/g

and, Hmax = vo^2*sin^2(theta)/g

so, as theta value changes T and Hmax also changes.

If theta values is given we can find T and Hamx.