Atoy rocket is launched from a 2.8 m high platform in such a way that its height, h (in meters), after t seconds is given by the equation h equals negative 4.9 t squared plus 18.9 t plus 2.8. how long will it take for the rocket to hit the ground?

3.98 s (approximately 4 s for practical purposes)

Explanation:

1) Equation given:

h(t) = - 4.9 t² + 18.9 t + 2.8

2) The time it will take the rocket fo hit the ground is given doing h (t) = 0

- 4.9 t² + 18.9 t + 2.8 = 0

3) You can solve that equation using the quadratic equation or working the equation to complete squares.

To use the quadratic equation you have a = - 4.9, b = 18.8, and c = 2.8

With that, t = [-18.8 +/- √ ( 18.8² - 4(-4.9)(2.8) ) ] / (2(-4.9) ) = 3.98 s

(I only picked the positive result because the negative does not have physical meaning).

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.