Question

A ball thrown is cought by the thrower after 4s how high did it go and with what velocity was it thrown? How far below the heighest point was the ball 3s after start? (g=9.80 ms-2)

Answer 1

Explanation:

I AM USING G=10

velocity:final V is 0 as the ball comes to rest

V=U+AT

0=u-10*4

u=40m/s

h=u^2/2g

=40*40/2*10

=80m

h=ut-1/2gt^2

=40*3-1/2 *10*3^2

=120-45

=75m from the ground

Or

5m from highest pt.

Answer 2

$\sf\star\bold\red{\underline{{QUESTION:-}}}$

✓✓A ball thrown is cought by the thrower after 4s how high did it go and with what velocity was it thrown? How far below the heighest point was the ball 3s after start? (g=9.8m/s²)

$\sf\star\bold\orange{\underline{{GIVEN:-}}}$

Time taken by the ball to reach the point $\sf\:=\dfrac{4}{2}$

$\sf\:t=2 sec$

$\sf\boxed\star\pink{\underline{\underline{{For\:upward\:motion:-}}}}$

• $\sf\:v=0m/s$

• $\sf\:g=-9.8m/{s}^{2}$

• $\sf\:t=2sec$

• $\sf\:u=\:?$

$\sf\star\bold\purple{\underline{{SOLUTION:-}}}$

$\sf\therefore\:v=u+gt$

$\sf\implies\:0=u+(-9.8)\times\:2$

$\sf\implies\:19.6m/s=u$

$\sf\blue{{\therefore\:u=19.6m}}$

then,

$\sf\boxed\star\red{\underline{\underline{{maximum\: height\: attained\:by\:the\:ball,}}}}$

$\sf\:s=ut+\dfrac{1}{2}g{t}^{2}$

$\sf\implies\:s=19.6\times\:2+\dfrac{1}{2}\times\:(-9.8)\times\:{2}^{2}$

$\sf\implies\:s=39.2+\dfrac{1}{2}\times\:(-9.8)\times\:4$

$\sf\implies\:s=39.2-19.6$

$\sf\green{{\implies\:s=19.6m}}$

Distance of the ball from the highest point is 3 sec after it was thrown = Distance covered by the ball in 1 sec from the highest point.

$\sf\:s=ut+\dfrac{1}{2}g{t}^{2}$

$\sf\implies\:s=0\times\:1+\dfrac{1}{2}\times\:9.8\times\:{1}^{2}$

$\sf\pink{{\implies\:s=4.9m}}$