Beatrice throws the 1st calculator straight up from the edge of the roof of a building. She drops a 2nd calculator off the side of the building 2 seconds after she has thrown the first one. If the height of the building is 40m, what must the initial upwards velocity of the 1st calculator be so the two calculators to hit the ground at exactly the same time?
first calc: hf=hi+vi*t-4.9t^2
second calc: hf=hi-4.9(t-2)^2
0=40+vi*t-4.9t^2
0=40-4.9(t-2)^2 well, two equations, two unknowns. start the with the second, solve for time, then put that into the first equation and solve for vi.
To solve this problem, we need to consider the motion of each calculator separately.
Let's first consider the motion of the calculator that is thrown straight up by Beatrice. We can use the equation of motion for vertical motion under constant acceleration:
h = ut + (1/2)at^2
Where:
h: height
u: initial velocity
t: time
a: acceleration
In this case, the height is 40m, the initial velocity will be upwards for the calculator thrown by Beatrice, and the acceleration is due to gravity (-9.8 m/s^2) acting downwards.
Using the equation, we can rearrange it to solve for the initial velocity (u):
u = (h - (1/2)at^2) / t
Now, let's consider the motion of the second calculator that is dropped off the side of the building. Since it's dropped, its initial velocity is zero. The only force acting on it is gravity, which gives it a constant downward acceleration of -9.8 m/s^2.
Since the second calculator is dropped 2 seconds after the first one is thrown, we need to calculate the height it has fallen during those 2 seconds. We can use the equation:
h = (1/2)at^2
Substituting the values, we get:
h = (1/2)(-9.8)(2)^2 = -19.6 m
Since the height is negative, it means the second calculator has fallen 19.6m in those 2 seconds.
Now, let's find the time it takes for both calculators to hit the ground. The time it takes for an object to fall from a certain height (h) can be calculated using the equation:
h = (1/2)gt^2
Rearranging the equation, we get:
t = √(2h / g)
Using this equation, the time it takes for the second calculator to hit the ground is:
t = √(2(40) / 9.8) ≈ 2.03 seconds
Now, we can substitute this calculated time into the equation for the initial velocity of the first calculator:
u = (h - (1/2)at^2) / t
= (40 - (1/2)(-9.8)(2.03)^2) / 2.03
≈ 19.51 m/s
So, the initial upwards velocity of the first calculator, thrown by Beatrice, must be approximately 19.51 m/s for both calculators to hit the ground at exactly the same time.