To find the height of the skydiver 7 seconds after they step out of the airplane, we can assume that the only force acting on the skydiver is gravity. The equation that relates height, initial velocity, time, and acceleration (due to gravity) is given by:
h(t) = -16t^2 + vt + h0
Where:
- h(t) is the height of the skydiver at time t,
- 16 is half of the acceleration due to gravity (32 ft/s^2),
- t is the time in seconds,
- v is the initial velocity (which is zero in this case),
- h0 is the initial height (10,000 feet).
Plugging in the values into the equation:
h(7) = -16(7)^2 + 0(7) + 10,000
h(7) = -16(49) + 10,000
h(7) = -784 + 10,000
h(7) = 9,216 feet
Therefore, the height of the skydiver 7 seconds after they step out of the airplane is 9,216 feet.
A skydiver jumped out of an airplane at the height of 10,000 feet. Assuming the initial velocity is zero, find the height of the skydiver 7 seconds after they step out of the airplane. Write a model, h(t)
that represents the height of the skydiver from the ground t seconds after they jump out of the airplane.(1 point)
Responses
9,126 feet
9,126 feet
69,216 feet
69,216 feet
9,888 feet
9,888 feet
9,759.9 feet
1 answer