How do I solve for the maximum height of someone who jumps straight up leaving the ground at 6.0 m/s.
1 answer:
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The horizontal distance covered by the ball before hitting the water is 70.4 m
Explanation:
The motion of the ball is the motion of a projectile, so it consists of two independent motions:
- A uniform motion along the horizontal (x) direction
- A uniformly accelerated motion along the vertical (y) direction
We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:
where:
s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)
is the initial vertical velocity of the ball, which is given by
where
u = 23.5 m/s is the initial velocity
is the angle of projection
Substituting,
is the acceleration of gravity, downward
Substituting everything into the equation we get:
Solving the equation for t, we find the time of flight of the ball:
t = -0.94 s
t = 3.59 s
We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.
Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:
Therefore, the horizontal distance covered in a time t is
And substituting t = 3.59 s, we find
So, the horizontal distance covered by the ball before hitting the water is 70.4 m.
Learn more about projectile motion:
#LearnwithBrainly
Answer:
K = 588.3 N/m
Explanation:
From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):
Fe - Ff - W*sin(22) = 0 Replacing Fe = K*X and then solving for X:
By conservation of energy:
Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:
K = 588.3 N/m
