one of the main battery of cannons protecting the entrance to San Francisco Bay were able t horizontally propel 7.3 kg cannon balls with an average speed of 26.2m/s. if the height of the cannon emplacement were 112m above San Francisco Bay,
a. how far from the cannon's muzzle (end of cannon where cannon ball exits) would the cannon ball land?
b. at what would the cannon need to be set so as to land half the distance?
My answer: a. 124m . b. 151 degrees
Are my answers correct??
how long does it take to fall 112m?
4.9t^2 = 112
t = 4.78 seconds
horizontal distance traveled is
4.78s * 26.2m/s = 125.26m
Rounding could account for the difference in our answers. I assume your 151° fits the equation, but it seems a bit odd to be firing backwards. Look for an acute angle that will give the same answer.
Where is 4.9 come from?
I got it , thank you !
To find the answers to your questions, let's analyze the projectile motion of the cannonball.
a. How far from the cannon's muzzle would the cannonball land?
To determine the horizontal distance traveled by the cannonball, we need to use the equation for horizontal motion:
distance = velocity × time
We don't have the time in this case, but we can calculate it using the vertical motion of the cannonball. The equation for vertical motion is:
vertical displacement = initial vertical velocity × time + (0.5 × acceleration × time^2)
Assuming there is no vertical displacement, because the cannonball is landing at the same height it was launched from, we can rearrange the equation to solve for time:
time = √(2 × vertical displacement ÷ acceleration)
Using the given height (112m) as the vertical displacement and the acceleration due to gravity (9.8 m/s^2) as the acceleration, we can calculate the time it takes for the cannonball to land.
time = √(2 × 112m ÷ 9.8 m/s^2) = 4.24s (rounded to two decimal places)
Now, we can use the equation for horizontal motion to find the distance:
distance = velocity × time = 26.2 m/s × 4.24s = 111.03m (rounded to two decimal places)
Therefore, the cannonball would land approximately 111.03 meters from the cannon's muzzle.
b. At what angle would the cannon need to be set so as to land half the distance?
To find the launch angle, we can use the equation:
range = (initial velocity^2 × sin(2θ)) ÷ gravitational acceleration
First, let's find the range when the launch angle is 45 degrees (half of 90 degrees):
range = (26.2 m/s)^2 × sin(90°) ÷ 9.8 m/s^2 = 113.17m (rounded to two decimal places)
Now, we need to find the angle that will give us half the distance (56.59m):
56.59m = (26.2 m/s)^2 × sin(2θ) ÷ 9.8 m/s^2
To solve for θ, we can rearrange the equation as:
sin(2θ) = (56.59m × 9.8 m/s^2) ÷ (26.2 m/s)^2
sin(2θ) = 0.50075
2θ = sin^(-1)(0.50075)
θ = (sin^(-1)(0.50075)) / 2
Calculating this value gives us:
θ = 14.87° (rounded to two decimal places)
Therefore, the cannon would need to be set at an angle of approximately 14.87 degrees to land half the distance.
Your answers:
a. Your answer of 124m is close, but the correct answer is approximately 111.03m (rounded to two decimal places).
b. Your answer of 151 degrees is not correct. The cannon would need to be set at an angle of approximately 14.87 degrees (rounded to two decimal places) to land half the distance.