# Finding velocity for projectile given distances and launch angle

• aqryus
In summary: I don't think I've ever heard of that before. Are you sure you're doing this correctly? In summary, the cat reaches a height of 35m over a distance of 103m/s at an angle of 55 degrees from the ground.
aqryus
Homework Statement
A cat is shot out of a cannon over a 30m high wall at an angle of 55 degrees from ground level. With what speed does it have to be launched to make it approximately 5m lover the wall if the wall is 250m from the cannon?
Relevant Equations
t =d/v
d = (v) (t) + (0.5) (a) (t)^2
i used t=d/v for the x direction using the formula t=250/vcos55

and subbed that into the formula for the y direction

d = (v) (t) + (0.5) (a) (t)^2
35 = (vsin55) (250/vcos55) - (4.9) (250/vcos55)^2

canceled out the first two v

35 = sin55 (250/cos55) - (306250/v^2cos3025)
35 = 357 - (306250/v^2cos3025)
35 cos3025 (v^2) = 306250
v^2 = 306250/25cos3025
v^2 = 10669
v = 103 m/s

but the answer is 54m/s so this is way off and I'm really not sure what I'm doing wrong. I thought maybe I should use 500m for x distance because the cat reaches 35m at the wall, which is only half of the trajectory (?). Any help is appreciated thank you.

aqryus said:
Homework Statement:: A cat is shot out of a cannon over a 30m high wall at an angle of 55 degrees from ground level.
Oh noes! A cat? Are you sure it wasn't a car or a rat? Poor kitty!

PeroK, Mark44 and erobz
I can only speak for myself when I say that I could be more helpful if understood your numbers and how you got them. I am baffled by equations like
35 = sin55 (250/cos55) - (306250/v^2cos3025)
Do you really mean cos3025? Where did you get an angle equal to 3025 in , I assume, radians?

I strongly recommend that you work out the solution algebraically without using numbers to get an expression for ##v^2## in terms of the projection angle ##\theta##, the horizontal distance ##x##, the wall height ##h## and, of course, ##g##. Then substitute the numbers. This would help us figure out (a) whether you got the wrong expression and where you went wrong or (b) whether you made a mistake in substituting the numerical values.

berkeman and erobz
aqryus said:
35 = (vsin55) (250/vcos55) - (4.9) (250/vcos55)^2

canceled out the first two v

35 = sin55 (250/cos55) - (306250/v^2cos3025)
It doesn't work this way. ##(v\cos(55))^2## is ##v^2 (\cos(55))^2##, with the cosine factor commonly written as ##\cos^2(55)##. The idea is that you're supposed to calculate ##\cos(55)## and then square that number, not square 55 and take the cosine of 3025.

Really, fire a cat out of a cannon? That's weird...

berkeman and PeroK

## What is the formula for finding velocity of a projectile?

The formula for finding velocity of a projectile is v = √(g * d / sin(2θ)), where v is the velocity, g is the acceleration due to gravity, d is the distance traveled, and θ is the launch angle.

## How do I calculate the launch angle for a projectile?

The launch angle for a projectile can be calculated using the formula θ = 0.5 * arcsin(g * d / v^2), where θ is the launch angle, g is the acceleration due to gravity, d is the distance traveled, and v is the velocity.

## What units should I use for the distance and velocity in the formula?

The units for distance should be in meters (m) and the units for velocity should be in meters per second (m/s) for the formula to work correctly.

## Can I use this formula for any type of projectile motion?

Yes, this formula can be used for any type of projectile motion, as long as the initial launch angle and distance traveled are known.

## Do I need to take air resistance into account when using this formula?

No, this formula assumes that there is no air resistance affecting the projectile. If air resistance is present, the actual velocity of the projectile may differ from the calculated velocity.

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