a boy standing on the tray of a utility travelling at 5 ms throws aa ball vertically upwads with a velocity of 7.35ms (relative to himself) and catches it again at the same level. what horizontal distance does the ball move while it is in the air
Vo = 5 + i7.35 = 8.89m/s[55.8o]
Dx = Vo^2*sin(2A)/g
Dx = 8.89^2*sin(111.6)/9.8 = 7.5 m.
pliz give the full workind f the above questio
To find the horizontal distance traveled by the ball while it is in the air, we need to consider the horizontal motion of the ball.
First, let's analyze the horizontal motion of the ball. Since the boy is standing on a utility tray traveling at a constant velocity, we can assume that there is no horizontal acceleration acting on the ball. Therefore, the horizontal velocity of the ball will remain constant throughout its motion.
The horizontal velocity of the ball is the same as the horizontal velocity of the utility tray, which is 5 m/s.
Now, let's calculate the time it takes for the ball to reach its maximum height and then fall back down to the same level.
When the boy throws the ball vertically upwards with a velocity of 7.35 m/s, we can calculate the time it takes for the ball to reach its maximum height using the formula:
v = u + at
where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (7.35 m/s)
a = acceleration (effect of gravity = -9.8 m/s^2 for upward motion)
t = time taken
Rearranging the equation, we have:
t = (v - u) / a
t = (0 - 7.35) / -9.8
t ≈ 0.75 seconds
Since the ball takes the same amount of time to fall back down to the same level, the total time for the ball's motion is approximately 2 * 0.75 seconds = 1.5 seconds.
Now, we can find the horizontal distance traveled by the ball:
distance = velocity * time
distance = 5 m/s * 1.5 s
distance = 7.5 meters
Therefore, the horizontal distance traveled by the ball while it is in the air is approximately 7.5 meters.