Question

A 5 kg ball is thrown upward with a speed 10ms-1 (Take g = = 10ms-²?). (a) Calculate the maximum height attained by it. (b) Find the potential energy when it reaches the highest point. [Ans = 5m , 250joule ]​

Answer 1

Given :-

A 5 kg ball is thrown upward with a speed 10ms-1 (Take g  = 10ms-²?)

To Find :-

(a) Calculate the maximum height attained by it. (b) Find the potential energy when it reaches the highest point.

Solution :-

(a) Maximum height attained by it.

We know that

v² - u² = 2gh

Where

v = final velocity

u = initial velocity

g = acceleration due to gravity

h = height

(0)² - (10)² = -2(10)(h)

Note - As it has stop at the final point. Therefore final velocity is 0

0 - 100 = -20h

-100 = -20h

-100/-20 = h

100/20 = h

5 = h

Hence, height is 5 m

(b) Potential energy when it reaches the highest point

We know that

PE = mgh

Where

PE = potential energy

m = mass

a = acceleration due to gravity

h = height

PE = (5)(10)(5)

PE = 50 × 5

PE = 250 J

Answer 2

EXPLANATION:-

(a) Maximum height attained by it :-

Let's apply 3rd equation of motion i.e. v² - u² = 2gh

Here ,v=final velocity

u=initial velocity

g=acceleration due to gravity

h=height

So let's put the value in formula

==> (0)² - (10)² = -2(10)(h)

[Note:-Here v is 0 because after reaching at the top height it stopped for sometime ]

==>0 - 100 = -20h

==>-100 = -20h

==>-100/-20 = h

==>100/20 = h

==>5 = h

So height is 5 m

(b) potential energy when it reaches the highest point :-

P.E=MGH

Here, P.E=POTENTIAL ENERGY

G=ACCELERATION DUE TO GRAVITY

H=HEIGHT ATTAINED

Let's put the value in formula :-

P.E=5×10×5

P.E=50×5

P.E=250