Question

a ball is thrown up vertically in a straight line with a velocity of 26.4 m/s determine its height from starting point after 5 seconds (g=10m/S2)​

Answer 1

distance= 26.4×5-1/2×10×5²

=1320/10-125

=132-125

=7metres

Answer 2

Answer:

7m

Explanation:

Given :

Initial velocity,u => 26.4m/s

Time taken,t => 5s

Acceleration due to gravity,g=> -10m/s^2

Note :-

If a body is moving downward initially then its velocity is taken as negative and if initially body is moving upward then its velocity is taken as positive.

The acceleration is negative when going up because the speed is decreasing. The acceleration is negative when going down because it is moving in the negative direction.

To Find :

Height,h = ?

Solution :

According to the second law of motion

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

Let, h be the maximum height reached and acceleration due to gravity be g. The time taken is 5s.

So,

$\sf{}h=ut-\dfrac{1}{2}gt^2$

$\sf{}\implies h=26.4\times 5 -\dfrac{1}{2}\times 10 \times 5^2$

$\sf{}\implies h=26.4\times 5 -\dfrac{1}{2}\times 10 \times 25$

$\sf{}\implies h=132 -\dfrac{1}{2}\times 10\times 25$

$\sf{}\implies h=132 -125$

$\sf{}\therefore h=7m$

Therefore,maximum height attained is equal to 7m.