# If a ball is thrown into the air with a velocity of 10m/s, its height in meters t seconds later is given by y=10t-4.9t^2.?

Update:

Find the average velocity for the time period beginning when t=1.5 and lasting 0.5 seconds.

Relevance

At time t = 1.5 s, the height is 3.975 meters. At time t = 2.0 seconds, the height is 0.40 m. Average velocity = (change in height) / (change in time)= (-3.575 ) / (0.5) = -7.15 m/s.

• The following equation is used to determine the height of the ball at a specific time.

y = vi * t – ½ * a * t^2

vi is 10 m/s. In this equation, -½ * a is equal to -4.9. So, the acceleration is -9.8 m/s^2. We can use the following equation to determine the ball’s vertical velocity at a specific time.

vf = 10 – 9.8 * t

At 1.5 seconds, vf = 10 – 9.8 * 1.5 = -4.7 m/s

The negative sign means the ball is falling. According to the update, the acceleration lasts for 0.5 seconds. Let’s use the same equation to determine the ball’s velocity at 2 seconds.

vf = 10 – 9.8 * 2 = -9.6 m/s

Since the ball has a constant acceleration, we can use the following equation to determine its average velocity.

Average velocity = ½ * (vi + vf)

Average velocity = ½ * (-4.7 + -9.6) = -7.15 m/s

I hope this is helpful for you.

• Vav = ((10-g*1.5)+(10-g*2))/2 = (-5-10)/2 = -7.5 m/sec

• I assume it is thrown straight upwards.

y = 10t - 4.9t²

y(1.5) = 15 – 4.9(1.5)² = 3.975 m

y(2) = 20 – 4.9•4 = 0.4 m

average velocity is usually listed as Δy/Δt or

V = (3.975 – 0.4) / 1 = 3.575 m/s

• y = 10t - 4.9t²

Average velocity is the change in position over change in time:

v = (y(2.0) - y(1.5)) / (2.0 - 1.5)

v = ((20 - 19.6) - (15 - 11.025)) / (0.5)

v = (0.4 - 3.975) / 0.5

v = -7.15

The average velocity between t=1.5 and t=2.0 is -7.15 m/s.