Question

A ball or thrown up vertically returns to

the thrower after 6 sec Find the velocity
. with which it was
thrown up and also, calucate the
maximum height it
reaches and its position after 4 secs.


please answer with explanation I will mark as brainlist ​

Answer 1

Answer:

$22 |2 |. < 2 {2 {2 \frac{ { {0}^{?} }^{2} }{?} }^{2} }^{2} | | 1211.0. { { { {? \times \frac{?}{?} \times \frac{?}{?} \times \frac{?}{?} }^{?} }^{?} }^{2} }^{2} \times \frac{?}{?} 2 {250316 + - \times \div \div \times \times \frac{?}{?} }^{?}$

Answer 2

Given:

The ball returns to the ground after 6 seconds.

t = 6/2 = 3 s.

the ball is thrown upwards, the acceleration is negative.

(1)  Velocity for upward motion

v=u + at

0=u+(−10)×3                  

u=30  m/s      

(2) Maximum height it reaches

$\begin{array}{l}\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2} \\\mathrm{~h}=30 \times 3+\frac{1}{2}(-10) \times 3^{2} \\\mathrm{~h}=45 \mathrm{~m}\end{array}$

(4) After 4 second, it starts to fall down.  

Let the distance by which it fall in 1 s   be   d

$\begin{array}{l}\mathrm{d}=0+\frac{1}{2} \mathrm{at}^{\prime 2} \quad \text { where } \mathrm{t}^{\prime}=1 \mathrm{~s} \\\mathrm{~d}=\frac{1}{2} \times 10 \times(1)^{2}=5 \mathrm{~m}\end{array}$

∴ Its height above the ground, h  =45−5=40 m  

Hence after 4 s, the ball is at a height of 40 m above the ground.