Answer:
The horizontal distance traveled by the projectile is 15.23 m.
Explanation:
Given;
angle of projection, θ = 25⁰
initial velocity of the projectile, u = 15 m/s
time of flight, t = 1.12 s
The the travelling path of the object is calculated as the range of the projectile
Therefore, the horizontal distance traveled by the projectile is 15.23 m.
The answer is 12.36. hoped this helped!
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Answer:
mgh= 10 x 8 x 10
= 800
but you can try 10 x 8 x 4^-1 x 10
centripetal acceleration is given by formula
given that
now we have
so the ratationa frequency is given by
