Suppose the height of an object fired straight up from the level ground is given by s(t)=300t-4.9t where s is measured in meters and t is measured in seconds. how fast is the object moving up as it leaves the ground after 2 seconds? find the average velocity on the interval [1,3]. How low will it take to hit the ground? What is the velocity when it hits the ground?

Thank you so much :)

Fay/Grace/Alice --

You need to show some work on what you've posted. Otherwise, the math tutors will think you just want to copy the answers down.

not sure about calculus, but physics will solve this

4.9t should be 4.9 t^2 (free-fall equation)

gravitational acceleration (downward) is 9.8 m/s^2

300 is the launch velocity in m/s

A) after 2 s ... 300 - (2 * 9.8)

B) acceleration is uniform , so same as A)

C) time up equals time down
... 2 (300 / 9.8)

D) impact velocity equals launch velocity

To find the velocity of the object at any given time, we need to find the derivative of the height equation with respect to time (t).

1. Taking the derivative of the height equation s(t) = 300t - 4.9t^2:

ds(t) / dt = 300 - 9.8t

2. To find how fast the object is moving up when it leaves the ground after 2 seconds (t = 2):

ds(2) / dt = 300 - 9.8(2)
= 300 - 19.6
= 280.4 meters per second

So, the object is moving up at a speed of 280.4 meters per second after 2 seconds.

3. To find the average velocity on the interval [1,3], we need to find the average rate of change of the height over that interval:

Average velocity = Δs / Δt

Δs = s(3) - s(1)
= (300(3) - 4.9(3)^2) - (300(1) - 4.9(1)^2)
= (900 - 4.9(9)) - (300 - 4.9)
= (900 - 44.1) - (300 - 4.9)
= 855.9 - 295.1
= 560.8 meters

Δt = 3 - 1
= 2 seconds

Average velocity = 560.8 meters / 2 seconds
= 280.4 meters per second

So, the average velocity on the interval [1,3] is 280.4 meters per second.

4. To find how long it takes for the object to hit the ground, we need to find the time (t) when the height (s) is equal to zero.

Setting s(t) = 0:

0 = 300t - 4.9t^2

4.9t^2 = 300t

4.9t^2 - 300t = 0

t(4.9t - 300) = 0

t = 0 (not considered since it's the starting point)

Or

4.9t - 300 = 0

4.9t = 300

t = 300 / 4.9
≈ 61.22 seconds

So, it will take approximately 61.22 seconds for the object to hit the ground.

5. To find the velocity when it hits the ground, substitute the time (t) value into the derivative equation:

ds(t) / dt = 300 - 9.8t

ds(61.22) / dt = 300 - 9.8(61.22)
= 300 - 599.516
≈ -299.52 meters per second

The negative sign indicates that the velocity is directed downward when it hits the ground, and the magnitude is approximately 299.52 meters per second.

So, the velocity when the object hits the ground is approximately -299.52 meters per second (directed downward).

To find the velocity when the object leaves the ground after 2 seconds, we need to find the derivative of the height function with respect to time.

The given height function is s(t) = 300t - 4.9t^2.

Taking the derivative of s(t) with respect to t:

s'(t) = 300 - 9.8t.

To find the velocity when the object leaves the ground after 2 seconds, we evaluate s'(t) at t = 2:

s'(2) = 300 - 9.8(2)
= 300 - 19.6
= 280.4 m/s.

Therefore, the object is moving upwards at a velocity of 280.4 m/s as it leaves the ground after 2 seconds.

To find the average velocity on the interval [1,3], we need to use the average velocity formula:

Average Velocity = (Change in Distance) / (Change in Time).

First, we need to find the height at t = 1 and t = 3:

s(1) = 300(1) - 4.9(1)^2
= 295.1 m.

s(3) = 300(3) - 4.9(3)^2
= 854.7 m.

Next, we find the change in distance:

Change in Distance = s(3) - s(1)
= 854.7 - 295.1
= 559.6 m.

The change in time is:

Change in Time = 3 - 1
= 2 seconds.

Now, we calculate the average velocity:

Average Velocity = Change in Distance / Change in Time
= 559.6 / 2
= 279.8 m/s.

Therefore, the average velocity on the interval [1,3] is 279.8 m/s.

To find how long it takes for the object to hit the ground, we need to set s(t) = 0 and solve for t:

0 = 300t - 4.9t^2.

Rearranging the equation:

4.9t^2 - 300t = 0.

Factoring out t:

t(4.9t - 300) = 0.

Solving for t, we have:

t = 0 (this is the initial time),
t = 300 / 4.9.

We discard the t = 0 solution as it represents the initial time when the object is on the ground.

Hence, it takes approximately t = 61.2 seconds for the object to hit the ground.

To find the velocity when it hits the ground, we substitute t = 61.2 into s'(t):

s'(61.2) = 300 - 9.8(61.2)
= 300 - 599.76
= -299.76 m/s.

Therefore, the velocity when the object hits the ground is approximately -299.76 m/s, indicating that it is moving downward.