Question

. An object is thrown up with a velocity of 98 m/s. Find the maximum

   height it reaches and also find the total time taken to return to the  

   starting point.  

explanation need ​

Answer 1

Answer:

Here, u= $\frac{98m}{s}$, $t_{1}$ =7s. When body reaches maximum height, its velocity becomes $v_{1}$ = 0.

Maximum height reached can be obtained using:

$v_{2}$ = $u_{2}$ +2ah

Now considering initial velocity as $v_{1}$ =0, displacement as s = $\frac{u2}{2g}$

We can find time for the body to come back from the maximum height using:

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

$\frac{u2}{2g}$ = $\frac{1}{2}$$gt^{2}$

t = $\frac{u}{g}$

Thus, total time, t= $\frac{2u}{g}$ =20 s

Now the time it takes to reach point P from ground and to reach from point P to ground is same i.e. 7s each time.  

Thus, it takes 20s−7s=13s to come back to the same point P since its projection.

Answer 2

Answer:

The height of the ball is 490m and the time taken to return to the starting point is 20 seconds

Explanation:

To find:

Height the ball reaches(s) and the time taken to return to the starting point(t)

Solution:

a = -9.8m/s(The negative acceleration as the ball is going up)

Let the time taken by the ball be t.

By Newton's first equation of motion,

v²=u²+2as

0²=98²+2X(-9.8)s

0=9604-19.6Xs

19.6Xs=9604

s=9604/19.6

s=490m

The height of the ball going up is 490m

For the time taken by the ball going up,

Let the time taken by the ball going up be t1,

Using Newton's first equation of motion,

v=u+at

0=98+at1.......(v=0)

0=98+(-9.8)t1

0=98-9.8Xt1

9.8xt1=98

t1=98/9.8

=10 :. t1=10s

The time taken by the ball going up is 10 seconds

For the time taken by the ball going down,

Let the time taken by the ball going down be t2,

Using Newton's first equation of free fall,

v=gt

98=9.8Xt2

t2=98/9.8

=10 :.t2=10s

Total time taken by the ball going down is 10 seconds

Total time taken=t=t1+t2

=10+10

=20 s

:. Time taken by ball to return to the starting point is 20 seconds.