When we push a volleyball into water,it comes back on the water surface
2 answers:
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Answer: after 1.75 seconds
Explanation:
The only force acting on the ball is the gravitational force, so the acceleration will be:
a = -9.8 m/s^2
the velocity can be obtained by integrating over time:
v = -9.8m/s^2*t + v0
where v0 is the initial velocity; v0 = -7.95 m/s.
v = -9.8m/s^2*t - 7.95 m/s.
For the position we integrate again:
p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0
where p0 is the initial position: p0 = 29m
p = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m
Now we want to find the time such that the position is equal to zero:
0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m
Then we solve the Bhaskara's equation:
Then the solutions are:
t = (7.95 + 25.1)/(-9.8) = -3.37s
t = (7.95 - 25.1)/(-9.8) = 1.75s
We need the positive time, then the correct answer is 1.75s
Answer:
(a) The time the ball stays in the air is approximately 4.66 seconds
(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters
(c) The maximum altitude of the ball is approximately 26.62 m
Explanation:
The given parameters of the question are;
The initial velocity at which a baseball cannon fires balls, u = 29 m/s
The angle at which the cannon is tilted, θ = 52°
(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;
Where;
t = The time it takes the baseball to reach maximum height
2·t = The total time of flight = The time the ball stays in the air
θ = The angle at which the ball is tilted = 52°
g = The acceleration due to gravity ≈ 9.81 m/s²
u = The initial velocity = 29 m/s
Therefore, we have;
The time the ball stays in the air, 2·t ≈ 4.66 seconds
(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;
∴ The distance from the cannon at which the ball will hit the ground = R
The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters
(c) The maximum altitude of the ball is equal to the maximum height reached by the ball, H, which is given as follows;
Therefore, we have;
The maximum altitude of the ball ≈ 26.62 m