h = 0.5(9.81 )(0.133 s)^2
h= 0.0872 m
Explanation:
2.
This question is equivalent to that of a projectile launched at an angle of 30 ° to the horizontal with an initial velocity of 15 m/s .
The distance of the point on the ground where the ball bounces for the 1st time , from the point of projection is basically the range of the projectile motion.
You can use the direct formula for range which is
(u²sin2x)/g =[ {15²sin(2*15)}/10 ] m
= [ ( 225*0.5 )/10 ] m = ( 225/20 ) m = 11.25 m.
Here, u represents initial velocity of projection,
x represents angle of projection with the ground
and
g is the acceleration due to gravit
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