A girls throws a tennis ball straight into the air with a velocity of 64 feet/sec. If acceleration due to gravity is −32 ft/sec2, how many seconds after it leaves the girl's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.

2 secs

Explanation:

Step 1:

Data obtained from the question:

Initial Velocity (U) = 64 feet/sec

Final velocity (V) = 0

acceleration due to gravity (g) = −32

ft/sec2

Time (t) =?

Step 2:

Determination of the time taken to reach the highest point. This is illustrated below:

V = U + gt

Applying the above equation, we can easily obtain the time (t) as follow:

0 = 64 + ( −32 x t )

0 = 64 - 32t

Collect like terms

0 - 64 = - 32t

- 64 = - 32t

Divide both side by -32

t = - 64/-32

t = 2 secs

Therefore, the time taken to reach the highest point is 2 secs

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down