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Gravitation | Gravitation - Exercise 10.6

Question 7

A ball is thrown vertically upwards with a velocity of 49 m/s.

Calculate

(i) The maximum height to which it rises,

(ii) The total time it takes to return to the surface of the earth.

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Given data:

Initial velocity u = 49m/s

Final speed v at maximum height = 0

Acceleration due to earth gravity g = -9.8 m/s^2 (thus negative as ball is thrown up).

By the third equation of motion,

v^{2}=u^{2}-2 g s

Substitute all the values in the above equation

\begin{array}{l} 0=(49)^{2}-2 \times 9.8 \times s \\ S=\frac{(49)^{2}}{2 \times 9.8} \\ s=122.5 \mathrm{m} \end{array}

Total time T = Time to ascend (Ta) + Time to descend (Td)

V = u – gt

0 = 49 – 9.8 x Ta

Ta = (49/9.8) = 5s

Also, Td = 5s

Therefore T = Ta + Td

T = 5 + 5

T = 10s

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Our top 5% students will be awarded a special scholarship to Lido.

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Question 7

A ball is thrown vertically upwards with a velocity of 49 m/s.

Calculate

(i) The maximum height to which it rises,

(ii) The total time it takes to return to the surface of the earth.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given data:

Initial velocity u = 49m/s

Final speed v at maximum height = 0

Acceleration due to earth gravity g = -9.8 m/s^2 (thus negative as ball is thrown up).

By the third equation of motion,

v^{2}=u^{2}-2 g s

Substitute all the values in the above equation

\begin{array}{l} 0=(49)^{2}-2 \times 9.8 \times s \\ S=\frac{(49)^{2}}{2 \times 9.8} \\ s=122.5 \mathrm{m} \end{array}

Total time T = Time to ascend (Ta) + Time to descend (Td)

V = u – gt

0 = 49 – 9.8 x Ta

Ta = (49/9.8) = 5s

Also, Td = 5s

Therefore T = Ta + Td

T = 5 + 5

T = 10s

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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