a train is moving at a speed of 5m/s. a person standing on the train throws a ball upwards at 10m/s. at the maximum height of the ball the train stopped. what is the deceleration of the train and what is the distance between the ball and the person finally?
Assume g=10 m/s².
The ball upward motion
0=v₀-gt,
t=v₀/g=10/10=1 s.
The train motion for time 1 s
0= v-at,
a=v/t=5/1 = 5 m/s²,
the distance
s= v t -at²/2 =5•1 -5•1²/2 = 2.5 m
To find the deceleration of the train, we need to consider the initial velocity of the train, the final velocity of the train, and the time it took for the train to stop.
Given:
Initial velocity of the train (u) = 5 m/s
Final velocity of the train (v) = 0 m/s (since the train stopped)
Time taken for the train to stop (t) = ?
Acceleration (a) can be calculated using the formula:
a = (v - u) / t
Since v = 0 m/s and u = 5 m/s:
a = (0 - 5) / t
a = -5 / t
The deceleration of the train is -5 m/s², indicating a negative acceleration or a deceleration.
Now let's calculate the height reached by the ball and the distance between the ball and the person when the train has stopped.
The ball was thrown upwards at a speed of 10 m/s. In projectile motion, when an object is thrown upwards, its initial velocity (u) is considered positive, and the acceleration due to gravity (g) is considered negative (-9.8 m/s²).
Using the formula for maximum height (H):
H = (u²) / (2 * |g|)
H = (10²) / (2 * 9.8)
H ≈ 5.102 m
At the maximum height, the ball momentarily comes to rest before falling back down. Hence, the distance between the ball and the person finally is equal to the height reached by the ball, which is approximately 5.102 meters.