Question

A ball is thrown vertically upward at initial velocity 20 ms.1 from the roof of a bus travelling with constant velocity 30 ms-1 along a straight road. The ball returns to the thrower's hand after the bus has travelled​

Answer 1

Answer:

In order to solve this question we must first solve for how long the ball is in the air:

`v_2-v_1 = at -gt t=(v_2-v_1)/a`

`t=(0-20)/-9.81=2.04s`

At 2.04s the ball has reached its maximum height

Next we must calculate its maximum height:

`d=((v_2+v_1)/2)t=(10)(2.04)=20.4m`

After 2.04s the ball has travelled 20.4m vertically.

Answer 2

Answer:

The correct answer to the given question is 4.08 seconds.

Explanation:

Given, the initial velocity of the ball thrown vertically upwards,

u=20m/s

The constant velocity of the bus is 30m/s

According to Newton's first equation of motion,

v-u=at

a is the acceleration due to gravity which will be taken as negative as the motion of the ball is vertically upwards that is in a direction opposite to that of gravitational force, v is the final velocity and u is the initial velocity.

As the ball reaches the hand of the thrower it comes to rest. So the final velocity of the ball is 0.

v=0 m/s

u=20 m/s

a= -9.8 $m/s^{2}$

0-20= -9.8xt

t= -20/9.8

t=2.04 seconds

Time taken by the ball to travel vertically upwards is equal to the time taken by the ball to reach the thrower's hand.

Therefore total time taken= (2.04+2.04)seconds

Time=4.08 seconds.

Learn more about equations of motion by clicking on the links given below-

https://brainly.in/question/16614356?referrer=searchResults

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