Fay/Grace/Alice --
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Suppose the height of an object fired straight up from the level ground is given by s(t)=300t-4.9t where s is measured in meters and t is measured in seconds. how fast is the object moving up as it leaves the ground after 2 seconds? find the average velocity on the interval [1,3]. How low will it take to hit the ground? What is the velocity when it hits the ground?
Thank you so much :)
2 answers
not sure about calculus, but physics will solve this
4.9t should be 4.9 t^2 (free-fall equation)
gravitational acceleration (downward) is 9.8 m/s^2
300 is the launch velocity in m/s
A) after 2 s ... 300 - (2 * 9.8)
B) acceleration is uniform , so same as A)
C) time up equals time down
... 2 (300 / 9.8)
D) impact velocity equals launch velocity
4.9t should be 4.9 t^2 (free-fall equation)
gravitational acceleration (downward) is 9.8 m/s^2
300 is the launch velocity in m/s
A) after 2 s ... 300 - (2 * 9.8)
B) acceleration is uniform , so same as A)
C) time up equals time down
... 2 (300 / 9.8)
D) impact velocity equals launch velocity
2 answers