A 10.00 kg block is placed at the top of a long frictionless inclined plane angled at 37.9 degrees relative to the horizontal. T
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The figure of the problem is missing: find in attachment.
(a) 1.64 s
The ball follows a projectile motion path. The horizontal displacement is given by
where
is the initial speed
t is the time
is the angle below the horizontal
We can rewrite this equation as
(1)
The vertical displacement instead is given by
(2)
where
is the acceleration of gravity
Substituting (1) into (2),
We know that for t = time of flight, the horizontal displacement is
We also know that the vertical displacement is
Substituting everything into the equation, we can find the time of flight:
(b) 36.5 m/s
We can now find the initial speed directly by using the equation for the horizontal displacement:
where we have
x = 50.8 m
Substituting the time of flight,
t = 1.64 s
We find:
(c) 47.1 m/s at 48.8 degrees below the horizontal
As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to
The initial vertical velocity is instead
And it changes according to the equation
So at t = 1.64 s (when the ball hits the ground),
So the impact speed is:
While the direction is:
