Question

An object is thrown up in the air. At time t (measured in seconds) the height of the object is h(t)=50t−16t2 (measured in feet). Find the velocity of the object at the following times: t=1 and t=3. Interpret the sign of the velocity in each case..

Answer 1

To find the velocity of the object, we take the derivative of the height function with respect to time. The velocity at t=1 is 18 ft/s, indicating that the object is moving upwards. The velocity at t=3 is -46 ft/s, indicating that the object is moving downwards.

Step-by-step explanation:

To find the velocity of the object, we take the derivative of the height function with respect to time. Let's find the velocity at t=1 first.
Given that the height function is h(t) = 50t - 16t^2, we can take the derivative of h(t) with respect to t to find the velocity function: v(t) = d/dt(50t - 16t^2).
Calculating the derivative, we get v(t) = 50 - 32t.
Substituting t=1 into the velocity function, we get v(1) = 50 - 32(1) = 50 - 32 = 18 ft/s.
The positive velocity indicates that the object is moving upwards at t=1.

Now, let's find the velocity at t=3.
Substituting t=3 into the velocity function, we get v(3) = 50 - 32(3) = 50 - 96 = -46 ft/s.
The negative velocity indicates that the object is moving downwards at t=3.