Projectile Motion Golf Ball Question

[landerphysics]

I'm sure you get a lot of projectile motion questions on here, sorry bout this. I used the search feature but I couldn't find anything quite like this problem and it's giving me some issues. It just doesn't seem like they give enough information for solving!

Homework Statement

A golfer can hit a golf ball a horizontal distance of over 300m on a good drive. What maximum height would a 301.5m drive reach if it were launched at an angle of 25.0 degrees to the ground.

Theta = 25 degrees
Distance (x) = 301.5m
At max height, Velocity (y) = 0
Max height is at 1/2t

Homework Equations

D(x) = Vi*(cos25)*t
V(x) = Vi*(cos25)
D(y)= Vi*(sin25)*t - .5*g*t^2
V(y), f = Vi*(sin25) - g*t
(Vy, f) ^2 = (Vi)^2*(sin25)^2 - 2g*D(y)

The Attempt at a Solution

I tried this problem for at least 45 minutes, but each time I just ended up erasing everything because it wouldn't work out. I was trying to work out the initial velocity, but I kept getting 2 unknowns in all the equations! Thanks for helping me :)

 

[lippyka]

Since you don't have enough information to solve for the initial velocity Vi, you're going to need to use an equation that does not require it. The equation you need is D(y)= Vi*(sin25)*t - .5*g*t^2. Using this equation, you can plug in the known values of theta (25 degrees), the distance (301.5m) and the acceleration due to gravity (g = 9.8 m/s^2). You will then be able to solve for the time t that it takes for the golf ball to reach its maximum height. Once you have the time, you can then plug it back into the equation to find the maximum height that the golf ball reaches. Hope this helps!

 

[Moetasim]

I completely understand your frustration with this problem. Projectile motion can be tricky to grasp at first, but with some practice and understanding of the equations involved, you will be able to solve problems like this with ease.

First, let's break down the given information. We know that the golf ball is being launched at an angle of 25 degrees to the ground, and that it travels a horizontal distance of 301.5m. We are also given the maximum height of the golf ball, which we can assume is the highest point it reaches during its flight.

To solve this problem, we need to use the equations for horizontal and vertical motion separately. In the horizontal direction, we know that the golf ball travels a distance of 301.5m, and we are looking for the initial velocity (Vi). Using the equation D(x) = Vi*cosθ*t, we can rearrange it to solve for Vi: Vi = D(x)/cosθ. Plugging in the values, we get Vi = 301.5m/cos25 = 327.2 m/s.

Now, let's look at the vertical motion. We know that at the maximum height, the velocity in the y-direction (Vy) is 0. Using the equation V(y),f = Vy,i - g*t, we can solve for the time it takes for the ball to reach its maximum height: t = Vy,i/g. Plugging in the values, we get t = 0/g = 0 seconds.

Next, we can use the equation D(y) = Vy*t - 1/2*g*t^2 to solve for the maximum height. Plugging in our known values, we get D(y) = 0*t - 1/2*g*(0)^2 = 0m. This does not seem right, as we know the ball should reach a maximum height.

The reason for this discrepancy is that we are assuming the ball is launched from ground level, which is not the case in this problem. The ball is actually launched from a height of 301.5m, and we need to take that into account when solving for the maximum height. The correct equation to use is D(y) = Vy*t - 1/2*g*t^2 + h, where h is the initial height of the ball. Plugging in our values, we get D(y) = 0*t - 1/2*g*(0