A projectile is launched efrom ground level to the top of a cliffwhich is 195m a

Question

A projectile is launched efrom ground level to the top of a cliffwhich is 195m away and 155m high. If the projectile lands on top ofthe cliff 7.6s after it is fired, find the initial velocity of theprojectile (magnitude and direction). Neglect air resistance.

My notes:

I have found a copy of the solution online, citing that the answeris 59 m/s, and 78o above the horizontal. This answeralso appears in the back of my textbook, in the section of answersto odd-numbered problems.

However, I disagree, and that the velocity should be 63.08 m/s,with the angle 66o above the horizontal.

My reason is that I believe the solution miscalculated the initial"x" velocity, and that the real initial x velocity should be 25.66m/s instead of the solution's proposed 12.5 m/s.

Can anyone verify my claims? Thanks!

Explanation / Answer

You are correct. The solution is 66 degrees and 63 m/s. ay=-9.8 (constant acceleration due to gravity) vy=-9.8t+v0*sin (velocity is theintegral of acceleration) hy=-4.9t2+v0*t*sin (positionis the integral of velocity) vx=v0*cos hx=v0*t*cos plug in the condition for the horizontal distance(hx=195 att=7.6): 195=v0*7.6*cos substitute for the initial velocity in the condition for thevertical distance (hy=155 at t=7.6) and solve for the angle: 155=-4.9*7.62+(195/(7.6*cos))*7.6*sin 155=-4.9*7.62+195*tan =66 degrees plug this back into either the horizontal or vertical distanceequations to solve for the initial velocity: v0=63 m/s it's worth noting though that 59m/s and 78 degrees is a solution tothe vertical "boundary condition" (ie. Hy(7.6)=155). howeverit fails the horizontal condition (Hx(7.6)=93), and thus is not thecorrect solution.

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