A related rates question?

Question

A ball is shot at an angle of 45 degrees into the air with initial velocity of 45 ft/sec. Assuming no air resistance, how high does it go? How far away does it land?

Can someone explain step by step how to do this problem.

thanks

Answer 1

Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time

Dx = v^2 / g

height = v^2 / (2g)

Hope this helps without actually solving it for you.

Answer 2

Step 1.Write down the given values:
Initial velocity =45ft/s
Ball shot at an angle of 45 degrees

Step 2. Solve for the vertical component of the velocity:
45sin45.
Step 3. Solve for the horizontal component of the velocity:
45cos45.
Step4. Use the formula v^2-u^2=2as to solve for the height. You know what the terms represent? v is the final velocity, u the initial velocity, a the acceleration of gravity=9.8*3.28ft/s^2, and s is the distance travelled.
Step 5. Take note that at its highest point the vertical component of the velocity of the ball is equal to zero. Also take note that the vertical component of the velocity when the ball was shot into the air is equal to the
vertical velocity when it falls to the ground. In using the above formula, imagine that you have a free falling body
starting from the top of the flight of the ball. This free falling body hits the ground with the same vertical velocity as when the ball was shot into the air. Knowing this it will be easier to get the values of v and u.
Step 6. Substitute known values in the above formula:

(45sin45)^2-0=2*9.8*3.28s
s=[45^2*0.707^2]/64.28
=15.7ft
Step 7. To solve for the range (how far away does it land?) you first solve for the total time it takes the ball to travel from the point where it was shot to the point where it lands.
Use the formula: s=ut+1/2at^2. Use the same data as in the above, but take note that the time t is only half the total time the ball takes to complete its flight. Think of the free falling body again, and you'll know why.
Step 8. Substitute known values:

51.6=0+1/2*9.8*3.28t^2
t^2=51.6/16.1
t=1.79s

Thus the total time of the flight is 1.79*2=3.58s.
Step 9. Use the basic formula: distance=velocity*time
The velocity to use here is the horizontal component of the velocity. Now substitute:

Distance (or Range)=45cos45*3.58
=114ft
Note : It will help a lot if you can draw the trajectory of the ball, and our imaginary free falling body.

Answer 3

always taking into consideration that there are Ux and Ux components of velocity you " break the problem in parts and examine it accordingly.
assuming that we examine the Y axis. with Uy known it is like throwing vertically a body at a known velocity (Uy) and we must know wha thte maximum heught will be.

the kinetic energy from the groung will be transformed to potential energy at the top. therfore,

1/2m*Uy^2 = m*g*h. ==>

Hmax = Uy^2 / 2*g => Hmax = 15.73ft

now you know the maximum height and the initial velocity and the distance traveled. you can trick this problem with two ways know i am showing the most clever one.

when it is at its maximum position it starts to fall. we know the height therefore we can figure out the time taken.

H = 1/2 g t^2 => t = sqrt (2s/g) = 0.9887
but this time is hlaf the time of all the projectile motion.
so total time on air will be 2*0.9887= 1.9774sec

we go back to the horizontala xis and we get
Max distance (S max ) realtes to Ux and total time traved on air.

so Ux = Smax / t =>

Smax = 45cos45 * 1.9774 = 62.92ft

Answer 4

u dont need related rates for this problem. use the formula for motion due to the influence of gravity. Find the vertical component of the initial velocity, and thats ur Vo. Find the time that it takes for it to hit the ground, then use that time and pug it into Dx=Vx(0)*t, to give u how far it went horizontally. Make sure u use the horizontal component of the initial velocity.