Selina Solutions Class 9 Physics Solutions for Exercise 3E in Chapter 3 - Chapter 3 Laws Of Motion

Question 44 Exercise 3E

A ball is thrown vertically upwards with an initial velocity of 49m/s. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g= $9.8m/s^2$ )

Answer:

Given: initial velocity u=49m/s, g= $9.8m/s^2$

(i) Assume ‘h’ to be the height, velocity = 0m/s at the greatest height

We know from the equation of motion,

$\begin{array}{l} v^{2}-u^{2}=2 g h \\ \Rightarrow 0-(49)^{2}=2(-9.8)(h) \\ \Rightarrow h=(49)^{2} / 19.6 \\ \Rightarrow h=122.5 m \end{array}$

(ii) Time taken before the ball reaches the ground is given by t=2u/g

t=2(49)/9.8

= 10s

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Exercises

Exercise 3A

Exercise 3B

Exercise 3C

Exercise 3D

Exercise 3E

Chapters

Chapter 1 – Measurements and Experimentation

Chapter 2 Motion In One Dimension

Chapter 3 Laws Of Motion

Chapter 4 Pressure In Fluids And Atmospheric Pressure

Chapter 5 Upthrust In Fluids Archimedes Principle And Floatation

Chapter 6 Heat And Energy

Chapter 7 Reflection Of Light

Chapter 8 Propogation Of Sound Waves

Chapter 9 Current Electricity

Chapter 10 Magnetism