Can Anyone Help?

[spudnick]

I have two problems that I need help with.

1) A baseball leaves a bat with a velocity of 35 m/s at an angle of 32 degrees. (A) What is the highest point in its path? (B) When will it strike a billboard in center field 8 m above the playing field.

Here's what I have for step (A):
Vf² - Vi²/2a= (0m/s)² - (35m/s)² / 19.6m/s²= -62.5 m

For step (B), would it just be: t=d/v= 8m/35m/s= .229 s ?

2) A rifle bullet is fired in a horizontal direction and strikes the bull's-eye 50m down range. The center of the bull's-eye is 100mm below the rifle's line of sight. What was the muzzle velocity of the bullet?

I don't even know where to begin for this problem.

If anyone can guide me in the right direction for these two problems it would be appreciated. Thank you!

[nautica]

Remember in your first question, you are only dealing with the Y direction when you are concerned with hieght. You must take sin of the angle. Then go from their.

Nautica

[ShawnD]

Originally posted by spudnick

1) A baseball leaves a bat with a velocity of 35 m/s at an angle of 32 degrees. (A) What is the highest point in its path? (B) When will it strike a billboard in center field 8 m above the playing field.


For part A, you have to make a formula that is specific to velocity in 1 direction - Y. Just completely ignore the X direction.
Vy = 35sin(32)
Now just fill in one of the formulas you already know.

V_f^2 = V_i^2 + 2ad

d = \frac{V_f^2 - V_i^2}{2a}

d = \frac{0 - (35sin(32))^2}{2(-9.81)}

d = 17.533m <------ highest point

For part B, just fill in that distance formula; using 35sin(32) as the velocity.

d = V_it + \frac{1}{2}at^2

I personally cannot answer this one without cheating and using my graphing calculator.... I can't figure out how to isolate t [g)]


2) A rifle bullet is fired in a horizontal direction and strikes the bull's-eye 50m down range. The center of the bull's-eye is 100mm below the rifle's line of sight. What was the muzzle velocity of the bullet?

What an intereting question.
Start off by finding the time it took for the bullet to drop that vertical distance. I could probably just use the distance formula but as said before, isolating t is tricky.

V_f^2 = Vi^2 + 2ad

V_f^2 = 2(9.81)(0.1)

Vf = 1.4007m/s

Now find the time it took to get that speed

V_f = V_i + at

t = \frac{V_f - V_i}{a}

t = \frac{1.4007}{9.81}

t = 0.14278s

Now find horizontal speed

v = \frac{d}{t}

v = \frac{50}{0.14278}

v = 350.1785m/s

That's about 1050ft/s which sounds reasonable for a muzzle velocity.

[HallsofIvy]

Originally posted by ShawnD
For part B, just fill in that distance formula; using 35sin(32) as the velocity.

d = V_it + \frac{1}{2}at^2

I personally cannot answer this one without cheating and using my graphing calculator.... I can't figure out how to isolate t [g)]


You don't know how to solve a quadratic equation?

d= 8= 35sin(32) t- 4.9t^2
8= 18.54 t- 4.9t^2
4.9 t^2- 18.54 t+ 8= 0

Use the quadratic formula.