)a toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. the rocket engine provides constant upward acceleration during the burn phase. at the instant of engine burnout, the rocket has risen to 64 m and acquired a velocity of 60 m/s. the rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. what is the time interval during which the rocket engine provides upward acceleration

The time interval during which the rocket engine provides upward acceleration is 2.1 s

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine burnout:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 = initial height

v0 = initial velocity

t = time

a = upward acceleration

g = acceleration due to gravity (downward)

The velocity of the rockey is given by this equation:

v = v0 + a · t     (v0 = 0 because the rocket is launched from rest)

v = a · t

and after burnout:

v = v0 + g · t

Where v = velocity at time t

We know that when the altitude is 64 m the velocity is 60 m/s. Then let´s use the following equation system:

y = y0 + v0 · t + 1/2 · a · t²    (y0 and v0 = 0)

v = a · t

Then:

64 m =  1/2 · a · t²

60 m/s = a · t

a = 60 m/s / t

Replacing "a = 60m/s / t" in the equation of height:

64 m = 1/2 ·( 60m/s / t) · t²

64 m = 30 m/s · t

t = 64 m / 30 m/s

t = 2.1 s

Then, the time interval during which the rocket engine provides upward acceleration is 2.1 s

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s