Solving the Displacement of a Catapulted Stone

[rover_dude]

Homework Statement

A stone is catapulted at time t = 0, with an initial velocity of magnitude 20.5 m/s and at an angle of 41.4° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.16 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.70 s.

I tried multiple solutions, but none of them are right, so I do not know what I am doing wrong.

The Attempt at a Solution

Ok so I think this is the closest I got to the answer, but I used the equation V=Vo(t)+(1/2)(a)(t)^2 So for the horizontal component at 1.16s I got x=20.5(1.16)+(1/2)(0)(1.16)^2 and I got x=23.78m. For the vertical component I got y=20.5(1.16)+(1/2)(-9.8)(1.16)^2 and y=17.18656m. For 1.7 seconds: x=20.5(1.7)+(1/2)(0)(1.7)^2 and I got x=34.85m for y: y=20.5(1.7)+(1/2)(-9.8)(1.7)^2 and y=20.689.
(I think these were the closest answers I got, but I tried different methods which were all wrong so help is appreciated)

 

[NascentOxygen]

So for the horizontal component at 1.16s I got x=20.5(1.16)+(1/2)(0)(1.16)^2

The horizontal component of initial velocity is not 20.5

For the vertical component I got y=20.5(1.16)+(1/2)(-9.8)(1.16)^2

The vertical component of initial velocity is not 20.5

 

[rover_dude]

So, If I have the correct initial velocity, this method would be correct?

 

[NascentOxygen]

So, If I have the correct initial velocity, this method would be correct?

Yes.

 

[rwisz]

Dear student,

Thank you for sharing your attempt at solving this problem. I can understand your frustration when your solutions do not match with the correct answers. Let's take a closer look at the problem and see if we can identify where you might have gone wrong.

First, it's important to always start with a clear understanding of the problem and the given information. In this case, we are launching a stone from a catapult with an initial velocity of 20.5 m/s and at an angle of 41.4° above the horizontal. We are asked to find the horizontal and vertical components of the stone's displacement at two different time points - 1.16 s and 1.70 s.

You correctly used the equation V=Vo(t)+(1/2)(a)(t)^2 to find the displacement at a given time. However, we need to be careful with the values we plug in for the initial velocity and acceleration. In this case, the initial velocity is not just 20.5 m/s, but rather 20.5 m/s at an angle of 41.4°. This means that the initial velocity has both horizontal and vertical components. The horizontal component can be found by multiplying the initial velocity by the cosine of the angle (20.5*cos41.4°), and the vertical component can be found by multiplying the initial velocity by the sine of the angle (20.5*sin41.4°).

Using these values, we can rewrite the equation for displacement as x=xo+Vox(t)+(1/2)(a)(t)^2 for the horizontal component, and y=yo+Voy(t)+(1/2)(a)(t)^2 for the vertical component. Note that xo and yo are the initial positions, which in this case are both 0 since the stone is launched from the origin.

Now, let's plug in the values for the horizontal component at t=1.16 s. We get x=0+20.5*cos41.4°*1.16+0=15.80 m. Similarly, for the vertical component, we get y=0+20.5*sin41.4°*1.16-9.8*(1.16)^2=-3.32 m. These values are different from what you calculated because you did not take into account the initial angle of launch.

For t=1.70 s, we get