A water rocket,launched from the ground,rises vertically with acceleration of 30 m/s² for 1.0 s when it runs out of "fuel." Disregarding air resistance,how high will the rocket rise? A)15 m B)31 m C)61 m D)120 m

Question

Quizplus · Accepted Answer

Using the kinematic equation:

h = ut + 1/2 at^2

where
u = initial vertical velocity = 0
a = acceleration = 30 m/s^2 (upward)
t = time = 1.0 s

Substituting the given values:

h = 0(1.0) + 1/2 (30) (1.0)^2
h = 15 m

Therefore, the rocket will rise 15 meters in the first second. However, since it has run out of fuel, it will continue to rise due to its initial velocity until it reaches the highest point of its trajectory. At that point, its velocity will be zero before it begins to fall back to the ground.

To find the maximum height reached by the rocket, we can use the fact that at the highest point of the trajectory, the final velocity will be zero. Therefore, using the kinematic equation:

v^2 = u^2 + 2as

where
v = final vertical velocity = 0
u = initial vertical velocity = 30 m/s (upward)
a = acceleration = -9.8 m/s^2 (downward, due to gravity)
s = displacement (which is equal to the maximum height reached)

Substituting the values:

0^2 = 30^2 + 2(-9.8) s
s = 61 meters

Therefore, the rocket will reach a maximum height of 61 meters.

The correct answer is choice C (61 m).

A Water Rocket,launched from the Ground,rises Vertically with Acceleration of 30