Answer:
We have that Assuming No air resistance,the initial horizontal velocity of the cannonball is
From the question we are told that
A cannonball is shot from a cannon at a launch angle of 35 and an initial velocity of 147 m/s
Generally the equation for the horizontal velocity is mathematically given as
Therefore
Therefore Assuming No air resistance,the initial horizontal velocity of the cannonball is For more information on this visit
Explanation:
Answer:
the speed of something in a given direction
Explanation:
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Answer:
The speed of man before he hits the ground is <u>23.35 m/s</u>
Explanation:
We know that:
Weight of Man - Force of Friction = Unbalanced Force
but, from Newton's 2nd Law of Motion:
unbalanced force = ma
Therefore,
W - F = ma
a = (W - F)/m
a = (mg - F)/m
where,
m = 81 kg
g = 9.8 m/s²
F = 103 N
a = [(81 kg)(9.8 m/s²) - 103 N]/81 kg
a = 8.52 m/s²
using 3rd equation of motion:
Vf² - Vi² = 2ah
here,
Vi = initial velocity = 0 m/s
Vf = Final Velocity before he hits ground = ?
Vf² - 0² = 2(8.52 m/s²)(32 m)
Vf = √545.28 m²/s²
<u>Vf = 23.35 m/s</u>
