Question 1013069
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The vertical and horizontal component of a projectile are 6m/s and 8m/s respectively take g = 10 m/s. Find


The initial velocity of the projectile
= sqrt(6^2 + 8^2) = 10 m/sec
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The greatest height reached by the projectile
h(t) = {{{(-10*t^2)/2 + 6t}}}
h'(t) = -10t + 6 = 0 at max
t = 0.6 second to get the greatest height. 
h(0.6) = 1.8 meters
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The time of  flight
= 2*tmax
= 1.2 seconds
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The range of the projectile.
= 1.2*8
= 9.6 meters