### Solution to problems – class 9 – Set 1 Q37, 38, 39, 40, 41

##### June 13, 2019

solution of 5 numericals from Q set 1 for class 9. Motion under gravity is the primary focus of the following problems.

Upward and downward movement, maximum height reached, time to reach the maximum height, velocity just before touching the ground etc are the touched upon points of these sort of numerical problems in physics. You can get the question set **here**

**icse
problems for class 9 physics – physics question **

**37) A piece of stone is thrown
vertically upwards. It reaches the maximum height in 3 seconds. If the
acceleration of the stone be 9.8 m/s^2 directed towards the ground, calculate
the initial velocity of the stone with which it is thrown upwards.**

Solution **Time to reach the max height = T=3 sec let the initial upward velocity = u**

**As per formula, T = u/g**

That means, u/g = 3

That means, u/g = 3

**So, u = 3 g = 3 x 9.8 = 29.4 m/s**

**38) A stone falls from a building and reaches the ground 2.5 seconds later. How high is the building ? ( g = 9.8 m/s^2)**

**icse
problems for class 9 physics – physics question **

Solution

During downward movement from rest, the initial velocity is u = 0**So the equation becomes: ****H = (1/2) g t ^{2} = (1/2) x 9.8 x 2.5^{2} = 30.6 m**

**39) A stone is dropped from a height of 20 m.( i) How long will it take to reach the ground?(ii) What will be its speed when it hits the ground? (g = 10 m/s^2)**

**icse
problems for class 9 physics – physics question **

Solution

H = 20 m

For downward movement from rest, U=0

To find out t, we will use the eqn, V = gt

so, t = V/g ….. (1)

**As yet we don’t know the value of V, to find it out we have to use another eqn. V ^{2} = 2 g H So, V^{2} = (2 x 9.8 x 20) V = 19.8 m/s**

So from eqn 1 we get t = V/g = 19.8/9.8 = 2 secs (approx) [ answer of i]

Final velocity as found above=19.8 m/s [answer of ii]

##### physics problem solution – motion under gravity

**40) A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall ? ( g = 9.8 m/s^2)**

Solution

The max height reached before it starts to fall = H = U^{2}/(2 g) = 20^{2}/(2 x 9.8)= 20.4 m

**icse
problems for class 9 physics – physics question **

**41) When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres.( a) What was the initial speed of the ball ?(b) How much time is taken by the ball to reach the highest point ? (g =10 m/ s^2)**

Solution

(a)At maximum height V =0

So from the eqn, V^{2} =U^{2} – 2gh

**We get, U = (2 g h) ^{1/2} = ( 2 x 10 x 5)^{1/2} = 10 m/s**

(b)The time required to reach the max height = t = U/g = 10/10 =1 sec