A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m/s2 for 6.0 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve?

h = 1094.69m

The maximum height above the ground the rocket will achieve is 1094.69m.

Explanation:

The maximum height h is;

h = height covered during acceleration plus height covered when the motor stops.

h = h1 + h2 ... 1

height covered during acceleration h1 can be derived using the equation of motion;

h1 = ut + 0.5at^2

Initial speed u = 0

h1 = 0.5at^2

acceleration a = 20 m/s^2

Time t = 6.0 s

h1 = 0.5 * (20 * 6^2)

h1 = 0.5 (20*36)

h1 = 360 m

height covered when the motor stops h2 can be derived using equation of motion;

h2 = ut + 0.5at^2 ... 2

Where;

a = g = acceleration due to gravity = - 9.8 m/s^2

The speed when the motor stops u;

u = at = 20 m/s^2 * 6.0 s = 120 m/s

Time t2 can be derived from;

v = u - gt

v = 0 (at maximum height velocity is zero)

u = gt

t = u/g

t = 120m/s / 9.8m/s^2

t = 12.24 seconds.

Substituting the values into equation 2;

h2 = 120 (12.24) - 0.5 (9.8*12.24^2)

h2 = 734.69376 m

h2 = 734.69 m

From equation 1;

h = h1 + h2. substituting the values;

h = 360m + 734.69m

h = 1094.69m

The maximum height above the ground the rocket will achieve is 1094.69m.