QUESTION: A projectile is fired with an inital speed of65.2m/s at an angle of 34

Question

QUESTION: A projectile is fired with an inital speed of65.2m/s at an angle of 34.5 degrees above the horizontal on alond flat firing range. Determine a) max height reached by theprojectile b) the total time in the air c)total horizontal distancecovered (that is the range) d)the velocity of the porjectile 1.50safter firing. (Giancoli P30) I need help begining the problem. Where would I begin and withwhat equation? Please be detailed in your answer. Can I find therange with the equationR=V02SIN/2. Explain why we shouldmodify it if we need to. Thank you. QUESTION: A projectile is fired with an inital speed of65.2m/s at an angle of 34.5 degrees above the horizontal on alond flat firing range. Determine a) max height reached by theprojectile b) the total time in the air c)total horizontal distancecovered (that is the range) d)the velocity of the porjectile 1.50safter firing. (Giancoli P30) I need help begining the problem. Where would I begin and withwhat equation? Please be detailed in your answer. Can I find therange with the equationR=V02SIN/2. Explain why we shouldmodify it if we need to. Thank you.

Explanation / Answer

initially vertical velocity of the projectile=65.2*sin34.5 vertical acceleration=g=9.8 let maximum height reached by the projectile=h at max. height velocity=0 v2=u2-2*g*h 0=(65.2*sin34.5)2-2*9.8*h h=69.58 m (ans a) let time taken to reach max height=t v=u-g*t 0=65.2*sin34.5-9.8*t t=3.77 s time taken to hit the ground from the max. height=T 69.58=u*t+(1/2)*g*T269.58=0+(1/2)*9.8*T2 T=3.77 s total time of flight=t+T=3.77+3.77=7.54 s (ans b) horizontal velocity=65.2*cos34.5 horizontal acceleration=0 let horizontal range=R R=u*t+(1/2)*g*t2R=u*t+0 R=65.2*cos34.5*7.54 R=405.15 m (ans c) horizontal velocity after 1.5 s=65.2*cos34.5=53.73 m/s let vertical velocity will be v v=u-g*t v=65.2*sin34.5-9.8*1.5 v=22.23 m/s velocity after 1.5s=[53.732+22.232]=58.15 m/s (ansd)

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