an object is dropped from a tower. the distance in t seconds is given by d(t)=4.9t^2
if the height of the tower is 146.9m , how fast is an object moving when it hits the ground??
V=dh/dt= d(4.9t^2/dt=9.8t
but t can be found from 146.9=4.9t^2
Your equation should have been
dt) = - 4.9t^2
To find out how fast an object is moving when it hits the ground, we need to determine its velocity at that moment. In order to do that, we'll need to find the derivative of the distance function, d(t).
The derivative of d(t) with respect to t will give us the velocity function, v(t). Using the power rule for differentiation, we can find v(t) as follows:
v(t) = d/dt (4.9t^2)
= 2 * 4.9 * t^(2-1)
= 9.8t
Now that we have the velocity function, we can determine the time it takes for the object to reach the ground. We know that when it hits the ground, the distance traveled is equal to the height of the tower, which is 146.9m. So, we can set the distance function equal to the height and solve for t:
d(t) = 146.9
4.9t^2 = 146.9
Dividing both sides of the equation by 4.9, we get:
t^2 = 30
Taking the square root of both sides, we find:
t = √30
Now we have the time it takes for the object to hit the ground, which is approximately 5.48 seconds. Finally, we can substitute this value into the velocity function to find the speed at that moment:
v(5.48) = 9.8 * 5.48
≈ 53.7 m/s
Therefore, when the object hits the ground, it is moving at a speed of approximately 53.7 m/s.