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March 27, 2017

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A home run is hit such a way that the baseball
just clears a wall 27 m high located 147 m
from home plate. The ball is hit at an angle
of 37◦ to the horizontal, and air resistance is
negligible. Assume the ball is hit at a height
of 1 m above the ground.
The acceleration of gravity is 9.8 m/s2 .
What is the initial speed of the ball?
Answer in units of m/s.
---------------------------------

I tried using these equations
xf=xi+vxit+.5g^2
tan theta= vy/vx => vy= vx*tan theta
yf=vyt+.5gt^2

147= 0+ vxi(4.30407) + 4.9(4.30407)^2=13.0638 wrong
I got t= 4.30407 by pluging in vy= vx*tan theta into yf=vyt+.5gt^2
1. The problem statement, all variables and given/known data

  • Physics- Mechaincs - ,

    147= 0+ vxi(4.30407) + 4.9(4.30407)^2
    well, I have issues with this. Why is gravity in the horizontal equation? (Besides,if it were, it would be negative).

    In the yf=vyt+.5gt^2 , the last term should be negative (-.5gt^2). Upwards is postive, downward negative.

  • Physics- Mechaincs - ,

    let V = initial velocity

    xf =xi + vxi t

    There is no acceleration in the horizontal direction, horizontal speed is constant!

    vxi = V cos 37
    so xf = 0 + .8 V t = 147
    solve for V t

    Now do y direction (up)
    yf = yi + V t sin 37 - .5 (9.8) t^2
    Note g is - if up is +
    27 = 1 + V t sin 37 -4.9 t^2
    we know V t
    solve for t

  • Physics- Mechaincs - ,

    and, on the yf, don't forget the initial height.
    Yf=yi+Vy*t-1/2 g t^2

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