A ball is dropped from the top of a 49.0 -high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 20.0 . The stone and ball collide part way up.
h1= g•t²/2
h2= v₀•t-g•t²/2
h1+h2 =h
g•t²/2 + v₀•t-g•t²/2 =49
t= 49/ v₀=49/20 =2.45 s.
h2= v₀•t-g•t²/2 =20•2.45 -9.8•2.45²/2=
=19.6 m
To determine where the ball and stone collide, we need to find the time it takes for each object to reach the same vertical position.
First, let's find the time it takes for the ball to reach this position. We can use the equation for free fall:
h = ut + (1/2)gt^2
Where:
h = height of the cliff (49.0 m)
u = initial velocity (0 m/s since it was dropped)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Rearranging the equation to solve for time (t), we have:
h = (1/2)gt^2
2h/g = t^2
sqrt(2h/g) = t
Substituting the given values, we can calculate the time it takes for the ball to reach the position where they collide.
Now let's find the time it takes for the stone to reach the same position. Since the stone is thrown straight up with an initial velocity of 20.0 m/s, we can use the kinematic equation:
v = u + gt
Where:
v = final velocity (0 m/s at the top)
u = initial velocity (20.0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Rearranging the equation to solve for time (t), we have:
v = u + gt
0 = 20.0 - 9.8t
9.8t = 20.0
t = 20.0 / 9.8
Substituting the given values, we can calculate the time it takes for the stone to reach the position where they collide.
Once we have both time values, we can compare them to determine when and where the ball and stone collide.