Motion in One Dimension Problem - 6

Description:

Question

A basketball player grabbing a rebound jumps 78.0 cm vertically. How much total time (ascent and descent) does the player spend?

a. In the top 15.0 cm of this jump and

b. In the bottom 15.0 cm?

Do your results explain why such players seem to hang in the air at the top of a jump?

Solution

For downward motion i.e. descent −

u = 0, S = 0.78 m, a = 9.8 ms^-2

By applying kinematic equation −

S = ut + 1/2 at²

Substituting the values −

0.78 = 0t + 1/2 × 9.8 × t²

0.78 = 4.9 × t²

t² = 0.78/4.9

t = √0.159 ≈ √0.16

t = 0.4 sec

Total time is given by −

Total time = t + t

Total time = 0.4 + 0.4 = 0.8 sec

a. For top 15 cm −

For descent −

u = 0, S = 0.15 m, a = 9.8 ms^-2

By applying kinematic equation −

S = ut + 1/2 at²

Substituting the values −

0.15 = 0t + 1/2 × 9.8 × t²

0.15 = 4.9 × t²

t² = 0.15/4.9

t = √0.0306

t = 0.175 sec

Total time is given by −

Total time = Time for descent + Time for ascent

Total time = 0.175 + 0.175 = 0.35 sec

b. For bottom 15 cm −

To calculate the time for bottom 15 cm, we will first calculate the time for 63cm (78cm – 15cm) from the top and then subtract it from the total time.

For 63 cm −

u = 0, S = 0.63 m, a = 9.8 ms^-2

By applying kinematic equation −

S = ut + 1/2 at²

Substituting the values −

0.63 = 0t + 1/2 × 9.8 × t²

0.63 = 4.9 × t²

t² = 0.63/4.9

t = √0.1285

t = 0.359 sec

Time for bottom 15cm is given by −

t = Time for 78cm - Time for 63cm

t = 0.4 - 0.36

t = 0.04 sec

Total time is given by −

Total time = Time for descent + Time for ascent

Total time = 0.04 + 0.04 = 0.08 sec

• Time taken in top is much larger than the time taken at the bottom, hence the player seems to hang-up in the air.