A bowling ball is dropped from a height of 40 feet.
a. Write a function that gives the height h (in feet) of the bowling ball after t seconds.
h(t)=
b. Find the height of the bowling ball after 1 second.
The height of the ball after 1 second is
feet.
c. How long does the bowling ball take to hit the ground? Round your answer to the nearest hundredth of a second.
The bowling ball takes about _ seconds to hit the ground.
Recognize that the word "dropped" suggests there is no initial velocity
so you equation must be
h(t) = -16t^2 + 40
b)
h(1) = .... , just sub in t = 1 into the h(t) equation
c) when it hits the ground, h=0
so
-16t^2 + 40 = 0
solve for t, using the positive result only
a. h(t) = 40 - 16t^2 (where t is the time in seconds)
b. To find the height of the bowling ball after 1 second, we substitute t = 1 into the equation:
h(1) = 40 - 16(1)^2
h(1) = 40 - 16(1)
h(1) = 40 - 16
h(1) = 24 feet
The height of the ball after 1 second is 24 feet.
c. To find how long it takes for the bowling ball to hit the ground, we set h(t) equal to 0 and solve for t:
0 = 40 - 16t^2
16t^2 = 40
t^2 = 40/16
t^2 = 2.5
t = √2.5
t ≈ 1.58 seconds
Therefore, the bowling ball takes about 1.58 seconds to hit the ground.
a. To find the function that gives the height h (in feet) of the bowling ball after t seconds, we can use the equation for the height of an object in free fall:
h(t) = initial height - (1/2) * g * t^2
where
- initial height is 40 feet (given),
- g is the acceleration due to gravity, which is approximately 32.2 feet per second squared.
Substituting these values into the equation:
h(t) = 40 - (1/2) * 32.2 * t^2
b. To find the height of the bowling ball after 1 second, we can substitute t = 1 into the function h(t):
h(1) = 40 - (1/2) * 32.2 * 1^2
h(1) = 40 - (1/2) * 32.2 * 1
h(1) = 40 - (1/2) * 32.2
h(1) ≈ 40 - 16.1
h(1) ≈ 23.9 feet
The height of the ball after 1 second is approximately 23.9 feet.
c. To find out how long it takes for the bowling ball to hit the ground, we need to find the time when the height is 0:
0 = 40 - (1/2) * 32.2 * t^2
Rearranging the equation:
(1/2) * 32.2 * t^2 = 40
32.2 * t^2 = 80
t^2 = 80 / 32.2
t^2 ≈ 2.484
Taking the square root of both sides:
t ≈ √2.484
t ≈ 1.574
Therefore, the bowling ball takes about 1.57 seconds to hit the ground (rounded to the nearest hundredth of a second).
a. To determine the height of the bowling ball at any given time, we can use the equation for free fall:
h(t) = h0 - (1/2)gt^2
Where:
h(t) = height of the ball at time t
h0 = initial height of the ball (40 feet in this case)
g = acceleration due to gravity (approximately 32.2 ft/s^2)
t = time in seconds
Applying this equation to the given scenario, we get:
h(t) = 40 - (1/2) * 32.2 * t^2
b. To find the height of the bowling ball after 1 second, we substitute t=1 into the equation:
h(1) = 40 - (1/2) * 32.2 * 1^2
= 40 - (1/2) * 32.2
= 40 - 16.1
= 23.9 feet
Therefore, the height of the ball after 1 second is 23.9 feet.
c. To determine how long it takes for the bowling ball to hit the ground, we need to find the time at which the height (h) reaches zero. Set h(t) = 0 and solve for t:
0 = 40 - (1/2) * 32.2 * t^2
Rearranging the equation:
(1/2) * 32.2 * t^2 = 40
32.2 * t^2 = 80
t^2 = 80 / 32.2
t^2 ≈ 2.48
Taking the square root of both sides:
t ≈ √2.48
t ≈ 1.57 seconds
Therefore, the bowling ball takes approximately 1.57 seconds to hit the ground.