A shingle falls straight down from the roof of a building that is 144 feet tall.
a. Use a vertical motion model to write an equation that gives the height h of the shingle t seconds after it falls.
b. When does the shingle hit the ground?
c. If the same shingle were dropped from a building that was twice as high, would it take twice as long to drop? Explain.
h=hi+vi*t+1/2 g t^2
h=144+0*t+1/2 (-9.8)t^2
b. when h=0, solve for t.
c. think. T=sqrt(hi/4.9)
a. To write the equation for the height of the shingle, we can use the vertical motion model. According to the model, the height of an object dropped from rest is given by the equation:
h(t) = h0 - (1/2)g*t^2
Where:
- h(t) is the height of the object at time t,
- h0 is the initial height (in this case, the height of the building),
- g is the acceleration due to gravity (approximately 32.2 ft/s^2),
- t is the time in seconds.
In this scenario, the initial height (h0) is 144 feet, so the equation becomes:
h(t) = 144 - (1/2) * 32.2 * t^2
b. To find when the shingle hits the ground, we need to determine the time (t) when the height (h(t)) becomes zero. We can set up the equation as follows:
0 = 144 - (1/2) * 32.2 * t^2
Solving this equation for t will give us the time when the shingle hits the ground.
c. If the same shingle were dropped from a building that was twice as high, it would not take twice as long to drop. According to the equation found in part (b), the time it takes for an object to hit the ground depends on the square root of the initial height. So, if the height is doubled, the time will only increase by a factor of the square root of 2, which is approximately 1.41. Therefore, the shingle would take approximately 1.41 times as long to drop from a building that is twice as high.