Question 1013069
The vertical and horizontal component of a projectile are 6m/s and 8m/s respectively take g=10m/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) = -10t^2 + 6t
h'(t) = -20t + 6 = 0 at max
t = 0.3 second
h(0.3) = 0.9 meters
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The time of  flight
= 2*tmax
= 0.6 seconds
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The range of the projectile.
= 0.6*8
= 4.8 meters