Asked by Melinda
An object is dropped from the observation deck of the Skylon Tower in Niagara Falls, Ontario. The distance, in metres, from the deck at t seconds is given by d(t)=4.9t^2
a) the height of the observation deck is 146.9 m. How fast is the object moving when it hits the ground?
The answer is 53.655 m/s.
Could someone explain how the book got 53.655 m/s.
a) the height of the observation deck is 146.9 m. How fast is the object moving when it hits the ground?
The answer is 53.655 m/s.
Could someone explain how the book got 53.655 m/s.
Answers
Answered by
Reiny
First you find the time it takes to hit the ground ...
4.9t^2 = 146.9
t^2 = 29.98 sec
t = √29.98 = 5.475
if d(t) = 4.9t^2
v(t) = 9.8t
v(5.475) = 9.8(5.475) = 53.659 m/s
Since you are in calculus, I am suprised that you did not know that the derivative of distance gives you velocity.
4.9t^2 = 146.9
t^2 = 29.98 sec
t = √29.98 = 5.475
if d(t) = 4.9t^2
v(t) = 9.8t
v(5.475) = 9.8(5.475) = 53.659 m/s
Since you are in calculus, I am suprised that you did not know that the derivative of distance gives you velocity.
Answered by
Colton Budz
It’s because this question is in unit one . We do not learn double derivatives until unit 3. So that is probably why. I had the same problem and my teacher wanted us to solve it with limits.
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