Question

A boy throws a ball in the air the ball reaches a height of 40 m. (10 marks )
How much time will it take the ball to come back?
If the ball reaches 60 m what is the initial velocity of the ball?
What will be final velocity before hitting the ground?
Draw a graph for both the conditions.​

Answer 1

Questions:

• 1) A boy throws a ball in the air the ball reaches a height of 40 m. How much time will it take the ball to come back.

• 2) If the ball reaches 60 m what is the initial velocity of the ball?What will be final velocity before hitting the ground? Draw a graph for both the conditions.

Solution:

1)

Max height reached=40m

We have to find time taken to reach the ground.

Since the ball reached max height of 40m, the velocity at 40m will be 0.

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,0){7}}\put(1,0.03){\framebox(1,7)}\multiput(0.3,0)(0.3,0){23}{\qbezier(0,0)(0,0)(-0.3,-0.3)}\put(3,6.9){\circle*{0.5}}\put(3,6.9){\vector(0,-1){1}}\put(3.5,6.1){ \bf u=0\ m/s }\put(0.3,3){\vector(0,-1){2.9}}\put(0.3,4.1){\vector(0,1){2.9}}\put(0.1,3.5){ \bf 40m }\end{picture}$

Use equation of motion:

$\boxed{s=ut+ \frac{1}{2}at^2}$

Here, u, the initial velocity is 0,

$\sf s= \frac{1}{2} at^2 \\\\ \sf 40= \frac{1}{2} \times 10 \times t^2 \\\\ \sf 40= \frac{1}{\cancel {2}} \times \cancel{10}\times t^2 \\\\ \sf 5t^2=40 \\\\ \sf t^2=8 \\\\ \sf t= 2 \sqrt{2} s$

2)

In this question, ball reaches 60 m height that is max height i.e v=0(final velocity =0) and since the ball is going upward, we will take g as -g

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,0){7}}\put(1,0.03){\framebox(1,7)}\multiput(0.3,0)(0.3,0){23}{\qbezier(0,0)(0,0)(-0.3,-0.3)}\put(3,0.5){\circle*{0.5}}\put(3,0.5){\vector(0,1){1}}\put(3.5,7){ \bf v=0\ m/s }\put(0.3,3){\vector(0,-1){2.9}}\put(0.3,4.1){\vector(0,1){2.9}}\put(0.1,3.5){ \bf 60m }\end{picture}$

We will use equation $\boxed{v^2=u^2+2as}$

$\sf 0^2= u^2-2gs \\\\ \sf u^2=2gs \\\\ \sf u^2= 2 \times 10 \times 60 \\\\ \sf u^2= 1200 m/s \\\\ \sf u= 20 \sqrt{3} m/s$

Since, the initial velocity and final velocity at same point is always same,the final velocity before reaching the ground will be equal to initial velocity i.e $20 \sqrt{3} m/s$

Tip: open in chrome desktop mode.

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