A ball is gently dropped from a height of 5m. If its velocity increases uniformly at the rate of
10m/s2 with what velocity will it strike the ground ? After what time will it strike the ground
Answers
Answer:
V= 10 m/s²
t= 1 sec
Explanation:
Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’.
Given parameters
Initial Velocity of the ball (u) = 0
Distance or height of fall (s) = 5 m
Downward acceleration (a) = 10 m s-2
As we know
2as = v2 – u2
v2 = 2as + u2
v2 = (2 x 10 x 5 ) + 0
v2 = 100
Final velocity of ball (v) = 10 ms-1
t = (v – u)/a
Time taken by the ball to strike (t) = (10 – 0)/10
t = 20/10
t = 1 seconds
The final velocity with which ball will strike the ground is (v) = 10 ms-1
The time it takes to strike the ground (t) = 1 seconds
Answer:
- Ball will strike the ground with velocity 10 m/s
- and, Ball will strike the ground after 1 sec.
Explanation:
Given:
- Ball is gently dropped from a height so, Initial velocity of ball, u = 0
- Height from which ball is thrown = 5 m
- Acceleration of ball, a = 10 m/s²
To find:
- Velocity with which ball will strike the ground, v =?
- Time taken by ball to reach the ground, t =?
Formulae required:
- First equation of motion
v = u + a t
- Third equation of motion
2 a s = v² - u²
[ Where v is final velocity, u is initial velocity, a is acceleration, t is time taken and s is distance covered ]
Solution:
In this case, height from which ball is thrown will be equal to the distance covered by the ball to reach the ground, so
- Distance covered by ball, s = 5 m
Now,
Using third equation of motion
→ 2 a s = v² - u²
→ 2 ( 10 ) ( 5 ) = v² - ( 0 )²
→ 100 = v²
→ v = 10 m/s
Using first equation of motion
→ v = u + a t
→ 10 = ( 0 ) + ( 10 ) t
→ 10 = 10 t
→ t = 1 sec
Therefore,
- Ball will strike the ground with velocity 10 m/s.
- and, Ball will strike the ground after 1 sec.