If a bowling ball is dropped out of an airplane, what will its speed (velocity) be after 3.4
seconds, assuming no air resistance? (Hint: Use
the equation v = 32t).

The velocity of the ball is 108.8 ft/s downwards.

Step-by-step explanation:

When the ball is dropped, the only force acting on the ball will be the gravitational force. Then the acceleration of the ball will be the gravitational acceleration, that is something like:

g = 32 ft/s^2

To get the velocity equation we need to integrate over time, to get:

v(t) = (32ft/s^2)*t + v0

where v0 is the initial velocity of the ball. (t = 0s is when the ball is dropped)

Because it is dropped, the initial velocity is equal to zero, then we get:

v(t) = (32ft/s^2)*t

Which is the same equation that we can see in the hint.

Now we want to find the velocity 3.4 seconds after the ball is dropped, then we just replace t by 3.4s, then we get:

v(3.4s) = (32ft/s^2)*3.4s = 108.8 ft/s

The velocity of the ball is 108.8 ft/s downwards.

b. t=-25h+3

step-by-step explanation:

its d. the first coordintaes are 10,32

step-by-step explanation: