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

Respuesta :

1. the max height is solved from v = square root of 2 gy where g is equal to 9.8 m/s2/ hence y is equal to 8.62 meters. 

 2. time at max height is solve when velocity = 0. ( 0 = -9.81*t +13); t = 1.325 seconds 

3. time required for the entire trip is 2*1.325 equal to 2.65 seconds. 

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

A. The maximum height reached by the ball is 8.62 m

B. The time taken to reach the maximum height is 1.33 s

C. The time required for the entire trip is 2.66 s

D. The ball's velocity just before it hits the ground upon returning is 13.034 m/s

A. How to determine the maximum height

  • Initial velocity (u) = 13 m/s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 13² – (2 × 9.8 × h)

0 = 169 – 19.6h

Collect like terms

0 – 169 = –19.6h

–169 = –19.6h

Divide both side by –19.6

h = –169 / –19.6

h = 8.62 m

B. How to determine the time

The time taken to reach the maximum height can be obtained as follow:

  • Initial velocity (u) = 13 m/s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Time to reach maximum height (t) =?

v = u – gt (since the ball is going against gravity)

0 = 13 – (9.8 × t)

0 = 13 – 9.8t

Collect like terms

0 – 13 = –9.8t

–13 = –9.8t

Divide both side by –9.8

t = –13 / –9.8

t = 1.33 s

C. How to determine the time for the entire trip

  • Time to reach maximum height (t) = 1.33 s
  • Time for the entire trip (T) =?

T = 2t

T = 2 × 1.33

T = 2.66 s

D. How to determine the velocity

The velocity of the ball before hitting the ground can be obtained as follow:

  • Time to reach ground from maximum height (t) = 1.33
  • Initial velocity (u) = 0 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 1.33)

v = 0 + 13.034

v = 13.034 m/s

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