A toy rocket has a mass 0f 350g at launch.
The force it produces is 15N and it is fired at
an angle of 65° to the horizontal. What is the
initial acceleration of rocket as a vector?
To determine the initial acceleration of the rocket as a vector, we need to resolve the force into its vertical and horizontal components.
First, let's find the vertical component of the force. The force exerted vertically is given by F_vert = F * sin(angle), where F is the force and angle is the angle it makes with the horizontal.
So, F_vert = 15N * sin(65°) = 15N * 0.9063 = 13.5945N (rounded to the nearest fourth decimal place).
Now, let's find the horizontal component of the force. The force exerted horizontally is given by F_horiz = F * cos(angle), where F is the force and angle is the angle it makes with the horizontal.
So, F_horiz = 15N * cos(65°) = 15N * 0.4226 = 6.339N (rounded to the nearest fourth decimal place).
Since we know the total mass of the rocket is 350g, we need to convert it to kilograms by dividing by 1000.
Mass = 350g / 1000 = 0.35kg.
Now, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
So, F = m * a.
Since we only have the vertical component, we will only consider the vertical acceleration (a_vert) for now.
Substituting the known values into the equation, we get:
13.5945N = 0.35kg * a_vert.
Solving for a_vert, we have:
a_vert = 13.5945N / 0.35kg = 38.8414 m/s^2 (rounded to the nearest fourth decimal place).
Therefore, the initial vertical acceleration of the rocket is 38.8414 m/s^2 upwards.
For the horizontal acceleration (a_horiz), there is no net force acting horizontally because the force exerted in that direction is balanced by the opposite force (resulting in zero acceleration). Thus, the initial horizontal acceleration of the rocket is 0 m/s^2.
As a result, the initial acceleration of the rocket as a vector is the combination of the vertical and horizontal components: 38.8414 m/s^2 upwards vertically and 0 m/s^2 horizontally.