Question

a student jumps from height of 4.9m with initial horizontal of 4.5 m/s and lands on ground. How much time student takes to reach ground.(g=9.8m/s)
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Answer 1

Answer:

Solution

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Correct option is C)

e= co-efficient of restitution=

4

3

h= initial height =4.9 m

v= velocity just before collision with ground

u= velocity after collision with ground=ev

v=

2gh

=

2×9.8×4.9

=9.8 m/s

u=ev=

4

3

×9.8 m/s

T= time to collide with the ground 2nd time=

g

u

+

g

u

=

g

2u

=2×

4

3

×

9.8

9.8

=

2

3

s

Answer 2

Student takes 1 sec to reach ground if student jumps from height of 4.9m with Initial horizontal speed of 4.5 m/s

Given:

• A student jumps from height of 4.9m
• Initial horizontal speed of 4.5 m/s lands on ground.
• g=9.8m/s²

To Find:

• Time student takes to reach ground

​

Solution:

• S = ut + (1/2)at²

Step 1:

Initial velocity is Horizontal  hence Initial vertical velocity is 0 m/s

Step 2:

Use formula S = ut + (1/2)at²  and substitute u= 0 m/s a = g = 9.8 m/s³ and S = 4.9 m

4.9 = 0.t + (1/2)(9.8)t²

Step 3:

Solve for t

4.9 = 0 + 4.9t²

4.9 = 4.9t²

1 = t²

1 = t  

Hence Student takes 1 sec to reach ground