Question

12. Make a plot of the acceleration of a ball that is thrown upward at 20 m/s subject to gravitation alone (no drag). Assume upward is the +y direction (and downward negative y).

13. Make a plot of the same ball's velocity.

14. Make a plot of the same ball's position.

15. At what time will the ball be back at its starting position?

16. At what time will the same ball be at max height?

17. What will the velocity in the y-direction be at max height?

18. Do any of these answers change if the ball additionally has a horizontal component to its initial velocity?

Answer 1

Initial velocity of the ball = 20 m/s

Here we take g = 9.8 m/s² = 10 m/s²

upward acceleration = - g = - 10m/s²

Downward acceleration = g = 10 m/s²

Upward motion

For upward motion acceleration = -10 m/s²

u = 20 m/s

When the ball reaches the maximum height , v = 0 m/s

Using first equation of motion

v = u + at

t = (v - u )/a =( 0-20) / -10 = 2 s

Then distance covered in 2s is calculated using third equation of motion

v² - u^{​​​​​​2} = 2aS

S = v²- u^{​​​​​​2} / 2a = (0² - 20²⁾ / (2×10) = 400 /20 = 20 m

Downward motion

Initial velocity = 0 m/s

S = 20 m

a = 10 m/s²

v² = u²+ 2aS

Substituting we get

v² = 0 + 2×10×20 = 400

v = 20 m/s

Hence upward and downward motions are equivalent. Hence time taken for downward motion is also 2 s.

12) Acceleration -time graph

13) velocity -time graph

At t= 0s, Velocity = 20 m/s

at t = 2s ball reaches maximum height, velocity = 0 m/s

at t = 4s the ball hit the ground with velocity ,v = 20 m/s

Since in the graph we took downward motion along negative Y axis, this velocity at 4s is -20 m/s.

14) position time graph

Position of the ball at t = 2s is 20 m.

Since the downward motion is taken along negative Y axis we need to consider that the downward motion starts from -20 m and reaches ground at 4s.

15) time for upward motion = 2s

time for downward motion = 2s

Total time = 4 s

16) ball reaches the maximum height when t = 2s

17) at maximum height , velocity = 0m/s

18) vertical velocity remains the same though the ball have an additional horizontal component.

hence none of the values calculated above will change. They will remain the same.

what difference actually happens was that the ball will have a horizontal displacement too.

that is it will not fall finally at the starting point.