Initial speed and max height of Geoff Capes' brick throw

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Homework Statement


In 1978, Geoff Capes of the United Kingdom won a competition for throwing 4.83 lb bricks; he threw a brick a distance of 44.99 m. Suppose the brick left Capes’ hand at an angle of 48.19o with respect to the horizontal and the bricks land at the same height he threw them from.

Find the initial speed of the brick.

If Capes threw the brick straight up with the speed found in (a), what is the maximum height the brick could achieve? (Ignore air resistance.)

2. Homework Equations
im kinda stuck here.

3. The Attempt at a Solution
i can't find an equation!
 

Discussion

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Although they'd be helpful, you don't need any odd equations. Just the standard ones will do:

vf=vi+at
x=vx*t
y=vy0*t + 1/2*a*t2

Call the initial speed "v", and write out the x and y displacements in terms of time. You know that the x displacement should be 44.99m and the y displacement should be 0, so you can eliminate the time and solve for v.

To solve part b, how long does it take an object traveling upwards at v to lose all of its speed? How far does the object travel during this time?
 
haha thes equations arn't what i would call standard. I am kinda confused as to what your trying to say.
 
We want to find the brick's initial speed, so call it v. The kinematic equations would read:
x=v*cos(theta)*t
y=v*sin(theta) - 1/2*g*t^2

You know that when at some time t, x=44.99 m and y=0, but you don't care about what that time is. So you can use those two equations to eliminate t and solve for v, which is what you're looking for.

BTW what kinematics equations are you used to?
 
ideasrule said:
We want to find the brick's initial speed, so call it v. The kinematic equations would read:
x=v*cos(theta)*t
y=v*sin(theta) - 1/2*g*t^2

You know that when at some time t, x=44.99 m and y=0, but you don't care about what that time is. So you can use those two equations to eliminate t and solve for v, which is what you're looking for.

BTW what kinematics equations are you used to?
okay i kinda see what your saying but were are you getting angles from??
m1v1i+m2v2i=m1v1f+m2v2f
m1vi1=m2vi2=(m1+m2)Vf
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2+1/2m2v2f^2
 
physicsgurl12 said:
okay i kinda see what your saying but were are you getting angles from??
m1v1i+m2v2i=m1v1f+m2v2f
m1vi1=m2vi2=(m1+m2)Vf
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2+1/2m2v2f^2

Are you sure you posted the right question? Those are conservation of momentum and conservation of energy equations, and have nothing to do with projectile motion.
 
ideasrule said:
Are you sure you posted the right question? Those are conservation of momentum and conservation of energy equations, and have nothing to do with projectile motion.

yes, i posted the right question, those are the only kinematic equations i know (atleast that i can remember at the moment)
 
physicsgurl12 said:
yes, i posted the right question, those are the only kinematic equations i know (atleast that i can remember at the moment)

They're not kinematic equations. They represent conservation of momentum and conservation of energy, and aren't related to the motion of a projectile in freefall.

Have you tried working with the equations I suggested?

x=v*cos(theta)*t
y=v*sin(theta) - 1/2*g*t^2

with the appropriate x and y values