Answer to Question #170820 in Classical Mechanics for Nema

Question

You throw a ball upwards with an initial velocity of 29 m/s. How high will the ball go before its velocity will be zero and it begins to fall and how much time will it take for the ball to reach that height?

Answer 1

Given,

Initial velocity of the ball in upward direction (u)=29 m/s

Let the time taken by the ball be t and maximum height attain by the ball be h.

When ball will reached at the maximum height then it's velocity becomes zero.

Now, applying the newton's first law of motion.

Let the gravitational acceleration be $(g)=9.8 m/s^2$

$v=u-gt$

Now, substituting the values,

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

$\Rightarrow t = \frac{29}{9.8}sec$

$t=2.95sec\simeq 3sec$

hence total time taken by the ball be $T= 2t = 6sec$

Now, for the maximum height,

$v^2= u^2-2gh$

Now, substituting the values,

$\Rightarrow h=\frac{u^2}{2g}$

$\Rightarrow$ $h=\frac{29^2}{2\times 9.8}m$

$\Rightarrow h =42.9m$