How Does Launch Angle Affect the Horizontal Distance in Shot Put?

[wbetting]

Homework Statement

Suppose that a shot putter can put a shot at the worldclass speed v0 = 14.00 m/s and at a height of 2.160 m. What horizontal distance would the shot travel if the launch angle θ0 is (a) 45.00° and (b) 41.00°? The answers indicate that the angle of 45°, which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

Homework Equations

Horizontal distance = 14 * cos θ * [(14 * sin θ ÷ 9.8) + (2.106 ÷ 4.9)^0.5]

The Attempt at a Solution

so i have been using the formula above and am not getting the right answer. I don't know what to do. wrong formula maybe?

 

[cepheid]

Rather than just trying to blindly apply formulae, let's think through the problem. In order to figure out the horizontal distance travelled, you need to figure out how much time the projectile spends in the air (so that you can multiply the horizontal speed by this time to get the distance).

So, how much time does the projectile spend in the air? It is the sum of two different times: 1) The time required to go up to the maximum height, and , 2) the time required to come back down to the ground from this maximum height.

How would you go about finding 1 and 2, knowing that the vertical acceleration is constant (it is due to gravity)?

As hinted at in the problem, the slightly tricky part here is that the starting height is 2.106 m, whereas the final height reached (upon hitting the ground on the way back down) is 0.00 m.

 

[wbetting]

so i tried to use d=1/2gt^2 and got .664 seconds for time up but when i time required to come back down d for final height is 0 so how can i do 0=1/2gt^2 without getting 0?

 

[cepheid]

You can't use d = 1/2gt^2 on the way up, because you don't know what d is. I.e. you don't know what maximum height above 2.106 m it reaches.

What you can do is to use what you know about speed vs time for constant acceleration. What *must* be true about the vertical speed of the object at the instant that it reaches its max height?

 

[asteriatic]

I can provide some guidance on how to approach this problem. First, it is important to understand the physical principles behind the motion of a projectile. The horizontal distance traveled by a projectile is dependent on its initial velocity, launch angle, and the height at which it is launched. In this case, we are given the initial velocity and launch height, but we also need to consider the height at which the shot lands.

To determine the horizontal distance traveled, we can use the formula for range of projectile motion: R = v0^2 * sin 2θ / g, where R is the horizontal distance, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. However, this formula assumes that the launch and landing heights are the same. Since they are not in this problem, we need to modify the formula.

One approach is to break the motion into two parts - the horizontal motion and the vertical motion. We can use the formula for horizontal distance traveled, which is R = v0 * cos θ * t, where t is the time of flight. To find the time of flight, we can use the formula for vertical displacement, which is h = v0 * sin θ * t - (1/2)gt^2, where h is the vertical displacement.

Using these equations, we can solve for the time of flight for each launch angle and then plug that into the formula for horizontal distance to determine the final answer. It is important to note that the launch angle that maximizes the horizontal distance may not be the same as the angle that maximizes the range of the projectile.

In conclusion, it is important to carefully consider all the factors at play in a physics problem and use the appropriate equations to find the solution. It is also important to double-check calculations and make sure the units are consistent throughout the problem. I hope this helps guide you in finding the correct solution.