Question

The engine of a toy rocket supplies an average acceleration of 38.0 m/s/s to the rocket for an interval of 0.80 s.

A) If the toy rocket is launched vertically, how high does it rise in this interval?

Option 1: 5.60 m
Option 2: 15.20 m
Option 3: 30.40 m
Option 4: 76.00 m

Answer 1

Using the kinematic equation for distance under constant acceleration, the toy rocket will rise to a height of 12.16 meters during the interval, making Option 2 the closest match after rounding.

Step-by-step explanation:

To determine how high the toy rocket will rise during the interval of acceleration, we can use the equation for the distance covered under constant acceleration (without initial velocity, since it's launched from rest):

d = (1/2) a t2.

Here, a is the acceleration (38.0 m/s2) and t is the time interval (0.80 s).

Substituting the given values into the equation:

d = (1/2) × 38.0 m/s2 × (0.80 s)2,

d = 19.0 m/s2 × 0.64 s2,

d = 12.16 m.

Therefore, the toy rocket will rise to a height of 12.16 meters during the interval. Looking at the provided options, none match exactly, suggesting a possible mistake in the options given or the need to round to the nearest option. With rounding, we would choose Option 2, 15.20 m, as it's the closest to our calculated height.