Question

A ball is thrown vertically upwards from the ground
with a speed of 25.2 ms^-1. How long does it take to
reach its highest point and how high does it rise?
(Take, g = 9.8 ms^-2)
​

Answer 1

Answer:

$\large\bold\red{h=32.4\:m}\\\\\large\bold\red{t=2.6\:s}$

Explanation:

Given,

A ball is thrown vertically upwards.

• Speed,$u = 25.2\:m{s}^{-1}$
• Value of$g = 9.8\:m{s}^{-2}$

Let,

• Maximum Height reached = h
• Time taken to reach maxm ht. = t

Now,

We know that,

$\large \boxed{ \bold \purple{h = \frac{ {u}^{2} }{2g} }}$

Therefore,

Putting the value,

We get,

$= > h = \frac{ {(25.2)}^{2} }{2 \times 9.8} \\ \\ = > h = \frac{635.04}{19.6} \\ \\ = > \bold{h = 32.4 \: m}$

And,

Also we know that,

$\large \boxed{ \bold \purple{t = \frac{u}{g} }}$

Therefore,

Putting the values,

We get,

$= > t = \frac{25.2}{9.8} \\ \\ = > \bold{ t = 2.6 \: s}$