Projectiles dealing with a bullet

[gotpink74]

1. Homework Statement
A shell is fired from the ground with an initial speed of 1.60*10^3m/s at an initial angle of 56° to the horizontal.
(a) Neglecting air resistance, find the shell's horizontal range.

(b) Find the amount of time the shell is in motion.

Homework Equations

vector triangle sin and cosine

The Attempt at a Solution

all of my answers have been wrong i did the triangle and got hyp 1600 opp side 1326.46 bottom 894.7 from there i don't know where to go please HELP

 

[gneill]

You'll need to use the kinematic equations for projectile motion (horizontal and vertical components).

 

[gotpink74]

i know that i just can't figure out the horiz and vertical components is
vertical
Vi:1600
thats all i know

 

[OliverENewell]

Would be better to:

1) Resolve the horizontal by doing Cos (theta) x initial velocity
2) v=u+at to get time in the air
3) Get horizontal component with trig like in 1)
4) distance bullet traveled is s = d / t rearranged to give d = s x t

That's all I can give you without giving you the answer on a plate :L

 

[asteriatic]

I understand the importance of accurately solving problems and using the correct equations and methods. In this case, it seems that you may have made a mistake in your calculations. Let's take a closer look at the problem and see if we can figure out where you went wrong.

First, we need to understand the given information. We know that the shell is fired with an initial speed of 1.60*10^3m/s at an initial angle of 56° to the horizontal. This means that the initial velocity of the shell can be broken down into two components: a horizontal component and a vertical component. The horizontal component can be found using the cosine function, and the vertical component can be found using the sine function.

(a) To find the shell's horizontal range, we need to find the distance the shell travels in the horizontal direction before hitting the ground. This can be found using the equation: R = v0*cos(theta)*t, where R is the horizontal range, v0 is the initial velocity, theta is the initial angle, and t is the time the shell is in motion.

Plugging in the given values, we get R = (1.60*10^3m/s)*cos(56°)*t. Since we are neglecting air resistance, we can assume that the time the shell is in motion is the same as the time it takes for the shell to reach the ground. Therefore, we can use the equation h = v0*sin(theta)*t - (1/2)gt^2, where h is the height of the shell, g is the acceleration due to gravity (9.8m/s^2), and t is the time the shell is in motion.

Solving for t, we get t = (2*h)/(v0*sin(theta)). Plugging this into our first equation, we get R = (1.60*10^3m/s)*cos(56°)*(2*h)/(v0*sin(theta)). Simplifying this, we get R = (1.60*10^3m/s)*cos(56°)*2*h*sin(theta)/v0. Since sin(theta)/v0 is equal to the time of flight (t), we can rewrite this as R = (1.60*10^3m/s)*cos(56°)*2*h*t. Plugging in the given value for h (0m), we get R = 0m