this is the ball being thrown ofver the house problem. boy 1 is 5.0 m on 1 side

Question

this is the ball being thrown ofver the house problem. boy 1 is 5.0 m on 1 side of the house, the house is 5 m and boy 2 is 5 m on the other side of the house. You catch and throw the ball 1 m from the ground. height to roof is 3.0. roof angle is 45 degrees. I can't get the correct answer and can't find where my work is wrong. Y = Vf^2 = 2*9.80m/s^2*4.5m = 9.39 m/s. Time = Vx= Vix + ayT with time 0.958163 multiplied by 2 for up and down time = 1.96133. X= vi +vT 15/1.91633 = 7.82748 m/s. using a^2 * b^2 = c^2 (9.39m/s)^2 + (7.82748 m/s)^2 = 149.441 sqrt = 12.22 a) 18.2 m/s b) 19.3 m/s c) 17.1 m/s d) 16.0 m/s Please show me what I'm missing or doing wrong. I need intial velocity

Explanation / Answer

R=v2sin2/g

H=v2sin2/2g

R=15, H=2

so dividing we get

=280

v=13.30 m/s

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