A stone is dropped from the top of a tower 400 m high and at the same time another stone is projected upward
vertically from the ground with a velocity of 100 ms −1 . Find where and when the two stones will meet.
x=Vot+1/2at^2
For downward motion Vo=0,
g=9.8m/s
h=Vot+1/2at^2
=0+1/2×9.8t^2
=4.9t^2
Forupward motion Vo=100
g=-9.8m/s
|||^ly
t=4s
From above,
h=4.9t^2
=4.9 (4)^2
=4.9(16)
78.4m
Therefore,
The two stones meet at 78.4m below the tp of the tower after 4seconds.
Be sure to read through the Related Questions below. Many of these questions are very similar, so when you find a formula for your particular assignment, use it.
S=1/2at^2
400-s=ut-1/2at^2
400-5t^2=100t-5t^2
t=4s
S=5t^2
S=80m from above
or 400-80=320m from below
To find where and when the two stones will meet, we need to find the time it takes for each stone to reach that point.
First, let's consider the stone dropped from the top of the tower. We can use the equation of motion:
h = ut + (1/2)gt^2
where:
h = height (400 m)
u = initial velocity (0 m/s as it is dropped)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken
Plugging in the values, we get:
400 = 0 * t + (1/2) * 9.8 * t^2
400 = 4.9t^2
t^2 = 400/4.9
t^2 = 81.632653
t = √81.632653
t ≈ 9.04 seconds
Now let's consider the stone projected upward from the ground with a velocity of 100 m/s. We can use the same equation of motion:
h = ut + (1/2)gt^2
where:
h = height (unknown)
u = initial velocity (100 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken (same as before, 9.04 seconds)
Plugging in the values, we get:
h = 100 * 9.04 + (1/2) * 9.8 * (9.04^2)
h ≈ 452.16 meters
Therefore, the stones will meet at a height of approximately 452.16 meters above the ground. The time it takes for them to meet is approximately 9.04 seconds.