A ball is thrown upward from the ground with an initial speed of 16.0 m/s; at the same instant, another ball is dropped from a building 12 m high. After how long will the balls be at the same height above the ground?
d1 + d2 = 12 m.
d1 = Vo*t + 0.5g*t^2
d2 = 0.5g*t^2
16*t - 4.9*t^2 + 4.9*t^2 = 12
16*t = 12
t = 0.75 s.
An object is thrown upwards at an initial speed of 19.6 m/s from a height of 40.0 m. Determine how long it will take for the object to reach the ground.
To determine the time at which the two balls will be at the same height above the ground, we need to find the time it takes for each ball to reach that height.
For the ball thrown upward, we can use the kinematic equation for vertical motion:
y = y0 + v0t - (1/2)gt^2
Where:
- y is the vertical position or height
- y0 is the initial height (which is 0 for the thrown ball)
- v0 is the initial velocity (16 m/s)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time
Setting y equal to 12 m (the height of the dropped ball), we can rearrange the equation to solve for t:
12 = 0 + 16t - (1/2)(9.8)(t^2)
Simplifying:
4.9t^2 - 16t + 12 = 0
Now, we have a quadratic equation. We can solve for t using either factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 4.9, b = -16, and c = 12. Plugging in the values:
t = (-(-16) ± √((-16)^2 - 4(4.9)(12))) / (2(4.9))
t = (16 ± √(256 - 235.2)) / 9.8
t = (16 ± √(20.8)) / 9.8
We get two possible solutions for t: t1 and t2. Since we are interested in the positive value, we take:
t1 = (16 + √(20.8)) / 9.8
Now we can calculate the time it takes for the dropped ball to fall from the building:
t2 = √(2h / g)
Where:
- h is the height of the building (12 m)
- g is the acceleration due to gravity (9.8 m/s^2)
Plugging in the values:
t2 = √(2(12) / 9.8)
t2 = √(24 / 9.8)
t2 = √2.4489
t2 ≈ 1.565 s
Since the ball was dropped, the time it takes to reach a height of 12 m is equal to t2.
Now, to find the time at which the two balls are at the same height, we compare t1 and t2. Take the smaller value:
t = min(t1, t2)
Let's calculate t:
t = min((16 + √(20.8)) / 9.8, 1.565)
After evaluating the expression, the time at which the two balls are at the same height above the ground is the resulting value of t.