Asked by Lindsay
                A cannon is fired from a cliff 190 m high downward at an angle of 38o with respect to the horizontal. If the muzzle velocity is 41 m/s, what is its speed (in m/s) when it hits the ground?
I keep getting 64.28 m/s as my answer, but it's not right. Could someone tell me exactly what I’m doing wrong?
            
            
        I keep getting 64.28 m/s as my answer, but it's not right. Could someone tell me exactly what I’m doing wrong?
Answers
                    Answered by
            bobpursley
            
    Final KE=1/2 m vi^2 + mgh
= 1/2 m 41^2 + m*9.8*190
1/2 m vf^2= 1/2 m( 3.55E5)
vf= 59.5m/s
    
= 1/2 m 41^2 + m*9.8*190
1/2 m vf^2= 1/2 m( 3.55E5)
vf= 59.5m/s
                    Answered by
            Lindsay
            
    Would 59.5 be the final answer?? B/c that is not right, either...
    
                    Answered by
            bobpursley
            
    Check my work.  The initial velocity is given as two significant digits, so the final should be in ... digits.
    
                    Answered by
            Lindsay
            
    I keep getting the same answer as you. 
This is what I did to get my original answer:
V^2 = sqrt[Vx^2 + Vy^2] where...
Vx = 41 m/s * cos 38
Vy at impact can be calculated using
Vy^2 = (41 sin 38)^2 + 2 g H
I got 55.56 as my Vy. I have a feeling this is where I got messed up...
    
This is what I did to get my original answer:
V^2 = sqrt[Vx^2 + Vy^2] where...
Vx = 41 m/s * cos 38
Vy at impact can be calculated using
Vy^2 = (41 sin 38)^2 + 2 g H
I got 55.56 as my Vy. I have a feeling this is where I got messed up...
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