Can someone draw this diagram for me because I must be getting the wrong picture

[smashbrohamme]

Determine the maximum horizontal distance that a projectile will have at a height of 60 m above its origin if it is fired at an angle at 70degrees from the horizontal with a velocity of 50m/s

 

[smashbrohamme]

here is my diagram.

 

[gneill]

How about this:

 

[zgozvrm]

gneill's diagram is the correct one.

Note that there are 2 horizontal distances such that the height of the projectile is at 60m.

 

[rwisz]

Unfortunately, as a scientist, I am not able to draw the diagram for you. However, I can provide you with the necessary information to determine the maximum horizontal distance of the projectile.

To determine the maximum horizontal distance, we can use the equation x = (v^2 sin 2θ) / g, where x is the maximum horizontal distance, v is the initial velocity, θ is the firing angle, and g is the acceleration due to gravity (9.8 m/s^2).

Plugging in the given values, we get x = (50^2 sin 2(70)) / 9.8 = 222.2 m. This means that the maximum horizontal distance the projectile will travel is approximately 222.2 meters.

I hope this helps clarify the concept for you. It may also be helpful to sketch out a rough diagram to visualize the trajectory of the projectile. Remember to always double check your calculations and units to ensure accuracy. Good luck!