A ball of mass 40 g is thrown upward. It reaches to the maximum height of 80 m. At what height its kinetic energy reduced by 40%?
Answers
Given:
A ball of mass 40 g is thrown upward. It reaches to the maximum height of 80 m.
To find:
Height at which KE reduced by 40%.
Calculation:
First of all, let initial velocity be u :
Now, initial KE :
Now, KE is reduced by 40%.
- So, final KE = 60% (Initial KE)
Now, required height will be :
At height 32 metres, KE will be reduced by 40%.
Given:
mass of ball m = 40g
maximum height Hmax = 80 m
To find:
height at which kinetic energy is reduced by 40% h =?
Explanation:
- A ball of mass 40 g is thrown upward & reaches the maximum height of 80 m.
i.e. m = 40 g
Hmax = displacement = S = 80 m
- Hence at maximum height, the velocity of a ball becomes zero.
i.e. final velocity v = 0 m/s
- First, we have to find the initial velocity with which a ball is thrown upward.
- By the kinematical law of equation
+ 2as
0 = + 2 (-9.81) (80)
- After solving above equation,
u = 39.6181 m/s
- The kinetic energy of ball is calculated as:
K.E = 0.5 × m
= 0.5 × 0.040 × 39.6181 (39.6181)
K.E = 31.3918 J
- Now, 40% of K.E = 0.40 × 31.3918 = 12.5567 J
12.5567 = 0.5 × m
⇒ v = 25.05 m/s
- from the law of the kinematical equation, we can find the height
+ 2as
(25.05)(25.05) = (39.6181)(39.6181) + 2(-9.81) s
s = 48.01 m
Hence at a height of 48.01 m, the kinetic energy reduced by 40% of a ball.