Explanation:
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- The velocity of the ball with strikes the ground.
- The time of the ball after which it strikes the ground.
We know that,
- The ball is dropped from the height of 30 m as stated in the question. Therefore, it will have a initial velocity of 0 m/s. The distance travelled by the ball will be 30 metres as the distance travelled by the ball is equal to the height of the tower and that is 30 metres.
So, by using the third equation of motion, we will find out the final velocity of the ball.
Therefore,
∴ Hence, the required final velocity of the ball with which it strikes the ground is 24.5 m/s. Since, we know that the initial velocity of the ball is 0 m/s and the acceleration of the ball is 10 m/s². So, by using the first equation of motion, we will find out the time of the ball after which it stikes the ground.
Therefore,
∴ Hence, the required time of the ball after which it strikes the ground is 2.45 seconds.
Answer:
speed for last lap is 247.89 km/h
Explanation:
given data
velocity v1 = 203 km/h
velocity v2 = 199 km/h
no of lap n = 10
to find out
average speed for last lap
solution
we consider here distance d for 1 lap
so in first 9 lap time taken is
t1 = distance / velocity v2
t1 = 9d / 199 ...............1
and
for 10 lap time is
t2 = 10d / 203 .............2
so from 1 and 2 equation time for last lap
last lap time t3 = t2 - t1
t3 = 10d / 203 - 9d / 199
t3 = 0.004034 d
so speed for last lap is
speed = distance / time
speed = d / 0.004034 d
speed = 247.89 km/h
so speed for last lap is 247.89 km/h
Answer:
You're going 3.2 meters per second.
Explanation: