To solve this problem, we need to find the equation of motion for both balls and set them equal to each other to determine the velocity of the second ball.
Let's first consider the motion of the first ball. Its initial velocity is 120 m/s, and it is acted upon by gravity (-9.8 m/s^2). The equation of motion for the first ball can be written as:
y1 = 120t - 4.9t^2
where y1 is the height of the first ball in meters and t is the time in seconds.
Now, let's consider the motion of the second ball. We are looking for the velocity that the second ball must have in order to pass the first ball at 100 meters from the ground. Therefore, for the second ball to catch up with the first ball, their heights must be equal. We can set up the equation for the height of the second ball as:
y2 = vt - 4.9t^2
where v is the velocity of the second ball and t is the time in seconds.
Since the two balls pass each other at a height of 100 meters, we can set y1 = y2 = 100 and solve for v.
100 = 120t - 4.9t^2
100 = vt - 4.9t^2
From the first equation, we can solve for t:
120t - 4.9t^2 = 100
4.9t^2 - 120t + 100 = 0
Solving this quadratic equation for t, we find t = 20 seconds.
Now, substitute t = 20 seconds back into the equation for the second ball:
100 = v(20) - 4.9(20)^2
100 = 20v - 1960
20v = 2060
v ≈ 103 m/s
Therefore, the velocity of the second ball must be approximately 105.89 m/s to pass the first ball at 100 meters from the ground.
A ball is thrown vertically into the air at 120 m/s. After 3 seconds , another ball is thrown vertically. What is the
velocitymust the second ball have to pass the first ball at 100 m from the ground? Ans. 105.89 m/s
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