Question

A ball is thrown vertically upward with a velocity of 19.6 m/s. Find

(i) The maximum height reached by the ball.

(ii) the time taken by the ball to reach the maximum height
answer above 40 words

Answer 1

u=19.6m/s.v=0.
a=-9.8m/s^2.
v^2-u^2=2as.
-19.6^2=2*9.8*s
s=19.6m.
v=u+at.
-19.6=-9.8*t
t=2s

Answer 2

Answer:

A ball is thrown vertically upward with a velocity of 19.6 m/s.

(i) The maximum height reached by the ball  =  $19.208\ m$

(ii) the time is taken by the ball to reach the maximum height $1.96\ s$

Explanation:

We can use the first and third equation of motion to solve this problem.

• According to the first equation of motion,

        $\mathrm{v}=\mathrm{u}+\mathrm{at}$            ...(1)

• According to the third equation of motion,

        $\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{aH}$     ...(2)

Where

u - Initial velocity of the object

v - Final velocity of the object

a - acceleration

t -  time taken

From the question,

$u = 19.6\ m/s$

$v =0$

$a = g =- 10\ m/s^2$  (since ball throw vertically upward)

Substitute all these values into equation (2)

$0-(19.6)^{2}=2 \times(-10) \times \mathrm{H}$

$H = \frac{19.6^{2} }{20} = 19.208 \ m$

Now consider equation(1),

$0=19.6-10 \mathrm{t}$

$t = 1.96 \ s$