A cannonball is fired with a velocity of 551 m/s. If the cannon is aimed 12° above the ground, how far away does the cannonball land? Ignore air resistance and use 9.8 m/s2 for acceleration due to gravity.
vertical problem first
Vi = 551 sin 12
v = Vi - 9.8 t
v = 0 at top
so
t = Vi/9.8 at top
total time in air = 2 t
= Vi/4.9
now horizontal:
u = 551 cos 12
distance = u (Vi/4.9)
To calculate how far the cannonball lands, we need to break down the initial velocity into horizontal and vertical components. The horizontal component will determine the distance traveled, while the vertical component accounts for the effect of gravity.
First, we need to find the initial horizontal velocity. This can be calculated using the formula:
V_horizontal = V_initial * cos(angle)
where:
V_horizontal: Initial horizontal velocity
V_initial: Initial velocity of the cannonball
angle: Angle of the cannonball (12° above the ground)
Let's substitute the values given into the equation:
V_horizontal = 551 m/s * cos(12°)
Next, calculate the initial vertical velocity using the formula:
V_vertical = V_initial * sin(angle)
where:
V_vertical: Initial vertical velocity
Substituting the values given:
V_vertical = 551 m/s * sin(12°)
Now, we can calculate the time it takes for the cannonball to land by using the vertical component of velocity. The equation for calculating the time of flight is:
t = (2 * V_vertical) / g
where:
t: Time of flight
V_vertical: Initial vertical velocity
g: Acceleration due to gravity (9.8 m/s²)
Substituting the values given:
t = (2 * V_vertical) / 9.8 m/s²
Once we have the time of flight, we can calculate the horizontal distance traveled by multiplying the horizontal velocity by the time of flight:
distance = V_horizontal * t
Substituting the values:
distance = (551 m/s * cos(12°)) * ((2 * (551 m/s * sin(12°))) / 9.8 m/s²)
Now, we can simplify the equation and calculate the distance.
Note: Remember to convert the angle from degrees to radians when using trigonometric functions.
Let me calculate the distance for you.