Finding initial projectile velocity with angle and a point in the path?

[ICanHasHelp]

Homework Statement

A ball is hit from the ground with an angle of 36 degrees above the horizontal. The ball hits a target 30m ahead and 4m above the ground. Find the initial velocity of the ball.

Homework Equations

V_avg = Δd/Δt
a_avg = Δv/Δt
Δd = v_oΔt + 1/2aΔt²

The Attempt at a Solution

Δx = 30m
Δy = 4m
V_ox = V_o * cos 36
V_oy = V_o * sin 36
Δx = v_oxΔt + 1/2a_x(Δt)²
Δx = (V_o * sin 36)Δt + 1/2a_x(Δt)²
Since a_x = 0 then
Δx = (V_o * sin 36)Δt


This question appeared on a quiz today and caught me totally by suprise. My teacher had never assigned homework questions like this ever and my textbook did not have any questions like it either. I've asked my teacher for help, but he didn't give me a clear answer. I've tried this question again at home for another two hours but I still couldn't get anything out of it. I know I have to find Δt, but I can't seem to isolate it. I would put all my attempts on the post, but I have already destroyed them in my fustration. Please help me! I'm losing my mind!

 

[w3390]

That is the correct equation for the x component of motion.

However, the reason your getting stuck is because you need another equation so that you can eliminate the unknown variable time.

This is where the y component of the motion will come in. Do this like you did to find the x component.

Remember: When the question asks for the initial velocity, it wants the velocity in the direction of launch, not a component of the velocity.

 

[ICanHasHelp]

I'm sorry I didn't initially post it, but I did find Δy, which was

Δy = VoyΔt - -4.9m/s^2

I also tried substituting Δx/Δt in place of Vox and Δy/Δt for Voy, but I still could solve for Δt because Δt was always dividing the direction.

 

[w3390]

So first of all, the equation is delta Y = (Vo_y)*t + (1/2)(-9.8)t^2.

It should not be a minus negative 4.9. You also left off the t^2.

Also, your equation for delta x should be cos(36) not sin(36).

 

[rwisz]

I understand your frustration with this question. It can be challenging to encounter a problem that you are not familiar with or have not been taught how to solve. However, as a scientist, it is important to approach problems with a logical and systematic approach. Let's break down the problem and see if we can find a solution.

First, we know that the ball is hit at an angle of 36 degrees above the horizontal. This means that the initial velocity of the ball can be represented as V0 = V0x + V0y, where V0x is the horizontal component and V0y is the vertical component. We can use trigonometric functions to determine these components. V0x = V0 * cos 36 and V0y = V0 * sin 36.

Next, we are given the distance and height of the target, which we can use to set up equations using the kinematic equations you have listed. However, we need to be careful in choosing which equations to use. Since we are looking for the initial velocity, we need to use equations that do not have time (Δt) as a variable. Let's use the equation Δy = V0yΔt + 1/2gt^2, where g is the acceleration due to gravity (9.8m/s^2). Plugging in our values, we get 4m = V0 * sin 36 * Δt + 1/2 * 9.8m/s^2 * Δt^2.

We also know that the ball travels a horizontal distance of 30m, so we can use the equation Δx = V0xΔt, which simplifies to 30m = V0 * cos 36 * Δt.

Now we have two equations with two unknowns (V0 and Δt). We can solve for Δt in the second equation and substitute it into the first equation to solve for V0. This will give us the initial velocity of the ball.

I hope this helps guide you towards finding a solution to this problem. Remember to approach problems systematically and use the appropriate equations. If you are still having trouble, don't hesitate to seek help from your teacher or a tutor. Keep up the good work in your studies!