A toy rocket is launched vertically from ground level at time t = 0 s.The rocket engine provides constant upward acceleration during the burn phase.At the instant of engine burnout,the rocket has risen to 49.0 m and acquired an upward velocity of 60.0 m/s.The rocket continues to rise with insignificant air resistance in unpowered flight,reaches maximum height,and falls back to the ground.The maximum height reached by the rocket is closest to
A)233 m.
B)221 m.
C)209 m.
D)244 m.
E)256 m.

Question

Quizplus · Accepted Answer

To find the maximum height reached by the rocket, we use the equation of motion under constant acceleration, $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m/s at the maximum height), $u$ is the initial velocity (60.0 m/s), $a$ is the acceleration due to gravity (-9.8 m/s$^2$, negative because it's acting opposite to the direction of motion), and $s$ is the displacement (the additional height gained after the engine burnout). Rearranging the equation for $s$, we get $s = \frac{v^2 - u^2}{2a}$. Substituting the given values, $s = \frac{0 - (60.0)^2}{2(-9.8)} = \frac{-3600}{-19.6} = 183.67 m$. This is the height gained after the engine burnout. The total height reached is the sum of the height at burnout (49.0 m) and the additional height gained (183.67 m), which equals approximately 232.67 m, which is closest to 233 m.

A Toy Rocket Is Launched Vertically from Ground Level at Time