A toy rocket is launched from the ground with a velocity of 80.0 m/s [11.0°]. How far
will the rocket travel horizontally before it hits the ground?
a. 2.44 x 102 m
b. 2.54 x 102 m
c. 2.64 x 102 m
d. 2.74 x 102 m
To find the horizontal distance the rocket will travel before hitting the ground, we can use the horizontal component of the initial velocity and the time it takes to hit the ground.
First, we need to find the horizontal component of the initial velocity:
vx = v * cos(theta)
vx = 80.0 m/s * cos(11.0°)
vx = 80.0 m/s * 0.9816
vx = 78.528 m/s
Now, we need to find the time it takes for the rocket to hit the ground:
We can use the following kinematic equation to find the time:
y = voy * t + 0.5 * a * t^2
0 = voy * t + 0.5 * (-9.8 m/s^2) * t^2
0 = 80.0 m/s * sin(11°) * t + 0.5 * (-9.8 m/s^2) * t^2
0 = 80.0 m/s * 0.1913 * t - 4.9t^2
0 = 15.302t - 4.9t^2
0 = 15.302 - 4.9t
t = 3.126 s
Now that we have the time, we can find the distance:
distance = velocity * time
distance = 78.528 m/s * 3.126 s
distance = 246.003 m
Therefore, the rocket will travel 2.46 x 10^2 m horizontally before hitting the ground, so the answer is:
b. 2.54 x 10^2 m