A projectile return to its original height -4.08 s after being launched. during

Question

A projectile return to its original height -4.08 s after being launched. during which time, it travels 76.2 m horizontally If air resistance can be neglected, what was the projectile's initial speed? At what angle with respect to the horizontal was the projectile originally fired? If a projectile was fired with the same initial velocity described in parts (a) and (b) on Mars (acceleration due to gravity of 3.71 m/s^2), how long would it take for it to return to its original height and bow far will it have gone horizontally in that time?

Explanation / Answer

a) As, R = u2 * sin(2theta)/g

             t = 2usin(theta)/g

=>    R/t2 = [u2 * sin(2theta)/g]/[4u2sin2(theta)/g2]

=>    R/t2 = [g]/[2tan(theta)]

=>    76.2/4.082 = [9.8]/[2tan(theta)]

=> theta = 46.95 degree

Thus, initial speed , v = 27.35 m/s

b) angle of launch , theta = 46.95 degree

c)   distance on Mars =   27.352 * sin(2 * 46.95)/3.71

                                    = 201.15 m

       Time on Mars =   2 * 27.35 * sin(46.95)/3.71

                               = 10.77 s

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