Sam tosses a ball horizontally off a footbridge at 3.2 m/s. How much time passes after he releases it until its speed doubles?
The vertical velocity component Vy would have to increase to sqrt3 times the initial horizontal velocity, Vx, so that V^2 = Vx^2 + Vy^2 = 4 Vx^2, and
V = 2 Vx
To achive that velocity, it must fall a distance H such that
M g H = (1/2) M Vy^2 = 3/2 M Vx^2
H = [3/(2g)]*Vx^2
i don't understand your work. please try to explain it more thoroughly.
To find the time it takes for the speed of the ball to double after being released horizontally, we can use the formula for speed:
Speed = Distance / Time
Since the ball is tossed horizontally, its initial vertical velocity is 0 m/s. Therefore, the speed of the ball equals only its horizontal velocity, which is 3.2 m/s.
Let's assume it takes 't' seconds for the speed to double. The speed after doubling would then be 2 times the initial speed, which is 3.2 m/s * 2 = 6.4 m/s.
Applying the formula for speed, we can rewrite it as:
6.4 m/s = Distance / t
Since the vertical displacement of the ball (distance) is influenced only by gravity, we know that it follows the formula:
Distance = (1/2) * g * t²
Here, 'g' is the acceleration due to gravity, which is approximately 9.8 m/s².
Plugging in the values, we get:
6.4 m/s = (1/2) * 9.8 m/s² * t²
Simplifying the equation further:
12.8 m/s = 4.9 m/s² * t²
Now, we can solve for 't' by rearranging the equation:
t² = 12.8 m/s / 4.9 m/s²
t² = 2.61224489796
Taking the square root of both sides:
t = √(2.61224489796)
t ≈ 1.6163 seconds
Therefore, it takes approximately 1.6163 seconds for the speed of the ball to double after being released horizontally.
To find the time it takes for the speed of the ball to double, we can use the equation derived from the constant acceleration formula:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, since the ball is moving horizontally, the vertical acceleration can be ignored. Therefore, the acceleration, a, is equal to 0.
The initial velocity, u, is given as 3.2 m/s.
The final velocity, v, is twice the initial velocity, which is 2 * 3.2 m/s = 6.4 m/s.
Using the equation v = u + at and rearranging to solve for t, we have:
t = (v - u) / a
Since a = 0, the equation simplifies to:
t = (v - u) / 0
However, division by zero is undefined, so the ball will never double its speed in this scenario.
Therefore, the time it takes for the speed of the ball to double is undefined.