A ball of mass 80g is thrown vertically upwards with a velocity of 20m/s
calculate
1)the max height, from the height which the ball is thrown
2) the total time of flight from the point it is released and caught at the same point.(assume g=10 ms-2)
2) 2 sec up and 2 sec down
1) ave vel = 10 m/s
10 m/s * 2 s = ?
To solve these questions, we can use the equations of motion for vertical motion. We can assume that the upward direction is positive and downward direction is negative.
1) To find the maximum height reached by the ball:
First, we need to find the time it takes for the ball to reach its maximum height. We can use the equation:
v = u + at
where:
v = final velocity (which is 0 m/s at the top)
u = initial velocity (20 m/s upwards)
a = acceleration (gravity, which is -10 m/s^2)
t = time taken to reach maximum height
Rearranging the equation, we have:
0 = 20 - 10t
Solving for t, we get:
t = 20/10 = 2 seconds
Now, we can use the formula for displacement in vertical motion:
s = ut + (1/2)at^2
where:
s = displacement (maximum height)
u = initial velocity (20 m/s upwards)
t = time (2 seconds)
a = acceleration (gravity, which is -10 m/s^2)
Plugging in the values, we have:
s = (20 * 2) + (1/2) * (-10) * (2^2)
s = 40 - 20
s = 20 meters
Therefore, the maximum height reached by the ball from the point it was thrown is 20 meters.
2) To find the total time of flight:
The time of flight is the total time from when the ball is thrown until it returns to the same point. Since the ball goes up and then returns down, the time taken to reach the maximum height is equal to the time taken to return to the starting point.
Thus, the total time of flight is 2 * t = 2 * 2 = 4 seconds.
Therefore, the total time of flight from the point the ball is released to when it is caught at the same point is 4 seconds.