A projectile is fired with an initial speed of 11.3 m/s at an angle of 60° above

Question

A projectile is fired with an initial speed of 11.3 m/s at an angle of 60° above the horizontal from on top of a 49 m high cliff. Determine the

a) Time to reach maximum height

b) Maximum height above the base of the cliff reached by the projectile (from ground level)

c) Total time in the air

d) Horizontal range of the projectile

Explanation / Answer

V = 11.3m.sec angle a = 60 V_verticel = V_v = 11.3 sin60 = 9.79 V_Horizontal = 11.3cos60 = 5.65 a) At max height v = 0 v = u +at 9.79/9.8 = t => t = 1sec b) max height be H H = ut + .5at^2 => H = 9.79*1 + .5*9.8 = 19.59m c) Total time n the air be T T = 1 + time to fall (19.59+49)m h = 68.59 h = ut+ .5at^2 => t = square root(68.59*2/9.8) 3.74 sec T =1+ 3.74 = 4.74 sec d) range = t*V_h = 4.74*5.65 = 26.78m

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