A ball is dropped from rest from a height of 20.0 m. One second later a second ball is thrown vertically downwards. If the two balls arrive on the ground at the same time, what must have been the initial velocity of the second ball?
What's the acceleration you're told to use in your class?
9.8m/s^2? 10m/s^2? 9.81m/s^2?
I'll go with 10 because I'm lazy, just substitute to whatever you guys are using.
The first ball (dropped from rest) will take 2 seconds to reach the ground :
20 = 5 x t^2 and you t=2
As such, the second ball will reach the ground in 1 second flat :
20 = Vinitial x t + 5 x t^2 where t=1
20 = Vinitial + 5
Answer : Vinitial = 15
Really, you just needed to use the equation you're given : x = x0 + v0*t + 1/2a*t^2
To find the initial velocity of the second ball, we can use the kinematic equation of motion for vertical motion:
y = ut + (1/2)at^2
where:
- y is the vertical displacement (distance) traveled by the ball
- u is the initial velocity of the ball
- t is the time taken for the ball to reach the ground
- a is the acceleration due to gravity (approximately -9.8 m/s^2)
Let's analyze the motion of each ball separately:
For the first ball:
- Initial vertical displacement, y1 = 20.0 m
- Initial velocity, u1 = 0 (since it was dropped from rest)
- Time, t1 = t (as both balls arrive on the ground at the same time)
- Acceleration due to gravity, a = -9.8 m/s^2
Putting these values into the equation for the first ball, we have:
20.0 = 0 × t + (1/2)(-9.8)(t)^2
20.0 = -4.9t^2
Simplifying this equation, we get:
t^2 = 20.0 / -4.9
t^2 = -4.08
Since time cannot be negative, there seems to be an error in the problem statement or the calculation. However, assuming the time is positive, we can continue solving the problem.
For the second ball:
- Initial vertical displacement, y2 = 0 (since it is thrown from the ground)
- Initial velocity, u2 = ? (what we want to find)
- Time, t2 = t (as both balls arrive on the ground at the same time)
- Acceleration due to gravity, a = -9.8 m/s^2
Putting these values into the equation for the second ball, we have:
0 = u2 × t + (1/2)(-9.8)(t)^2
0 = u2 × t - 4.9(t)^2
Since we want to find u2, let's rearrange the equation:
u2 × t = 4.9(t)^2
u2 = (4.9(t)^2) / t
u2 = 4.9t
Therefore, the initial velocity of the second ball must be 4.9t, where t is the time taken for the balls to reach the ground.
Note: Since the calculated value of t^2 is negative in this case, it suggests a problem with the problem statement or calculation. It is not physically possible for a value squared to be negative.