A rocket is launched vertically and returns its launch point in 7 seconds. a. What us it’s launch velocity?
b. How high did it go?
C. What is its velocity right before it hits ground?

A. 35 m/s

B. 61.25 m

C. 35 m/s

Explanation:

From the question given above, the following data were obtained:

Time of flight (T) = 7 s

Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Time of flight (T) = 7 s

Time (t) take to reach maximum height =?

T = 2t

7 = 2t

Divide both side by 2

t = 7/2

t = 3.5 s

A. Determination of the Lauch velocity.

Time (t) to reach maximum height = 3.5 s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Initial velocity (u) =?

v = u – gt (since the rocket is going against gravity)

0 = u – (10 × 3.5)

0 = u – 35

Collect like terms

0 + 35 = u

u = 35 m/s

Thus, the Lauch velocity is 35 m/s

B. Determination of maximum height reached by the rocket.

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Initial velocity (u) = 35 m/s

Maximum height (h) =?

v² = u² – 2gh (since the rocket is going against gravity)

0² = 35² – (2 × 10 × h)

0 = 1225 – 20h

Rearrange

20h = 1225

Divide both side by 20

h = 1225 / 20

h = 61.25 m

Thus, the maximum height reached by the rocket is 61.25 m

C. Determination of the velocity before hitting ground.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 10 m/s²

Time (t) to reach the ground from the maximum height = 3.5 s

Final velocity (v) =?

v = u + gt (since the rocket is moving towards gravity)

v = 0 + (3.5 × 10)

v = 0 + 35

v = 35 m/s

Thus, the velocity of the rocket before hitting ground is 35 m/s

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.