An orange is thrown straight up with an intial velocity of10m/s. A small gun is
ID: 1663712 • Letter: A
Question
An orange is thrown straight up with an intial velocity of10m/s. A small gun is positioned 1 m below the orange'sintial position. Given the intial velocity of a bullet gunfired is 30m/s. Our goal is to hit the oragne when it reachesthe highest point of it's trajectory. a) how long after the orange has been thrown should wefire? b) how far below the highest was the orange when we fired thebullet in part a. An orange is thrown straight up with an intial velocity of10m/s. A small gun is positioned 1 m below the orange'sintial position. Given the intial velocity of a bullet gunfired is 30m/s. Our goal is to hit the oragne when it reachesthe highest point of it's trajectory. a) how long after the orange has been thrown should wefire? b) how far below the highest was the orange when we fired thebullet in part a.Explanation / Answer
a. Height attained byorange is given by third equation of motion v2 = u2 + 2* g * h v = finalvelcity oforange = 0,u = initialvelocity = 10m/s, g = - 9.8m/s2 ( - ve for upwards motion) 0 = 102 + 2* ( -9.8) * h height h = 100/19.8 = 5.10 m and first equationis v = u+ g * t time t = (0- 10) / ( -9.8) = 1.02 s Distance travelled bybullet H = h +1 = 6.10 m Time taken by bullet can be calculatedusing second equation H = U *T + (1/2) * g *T2 6.10 = 30* T + 0.5 * ( - 9.8) *T2 4.9 *T2 - 30 *T + 6.10 = 0 Solving this quadraticequation T = 5.91 s, 0.21s, sincefirst is not acceptable T = 0.21 s To reach highest point at same time, the gunmust be firedafter t - T s,i.e 1.02 -0.21 = 0.81 s. b. Height reached by orangein 0.81 s h' = u *t' + (1/2) * g *t'2 = 10 *0.81 + 0.5 * (- 9.8) *0.812 height oforange h' = 4.885 m height oforange h' = 4.885 mRelated Questions
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