1. A basketball player passes the ball to his teammate at an angle of 40degrees above the horizontal. The ball is released 1.5m above the floor at a speed of 5m/s. Find:
a) the maximum height reached by the ball from the floor
b) the horizontal required to reach the maximum height
c) the horizontal distance the ball traveled during this time
2. A book slides off a horizontal table 1m high with a speed of 2m/s. Find:
a) the horizontal distance of the book from the table when it strikes the floor
b) the time of travel and the horizontal distance traveled if the initial speed is doubled.
1. Vo = 5 m/s @ 40 Deg.
Xo = 5 cos40 = 3.83 m/s.
Yo = 5 sin40 = 3.21 m/s.
a. Yf^2 - Yoo^2 + 2g*h,
h = (Yf^2 - Yo^2) / 2g,
h = 1.5 + (0 - (3.21)^2) / -19.6=2.03m.
b.
c.Tr = (Yf - Yo) / g,
Tr = (0 - 3.21) / -9.8 = 0.328 s. = Rise time or time to reach max. ht.
Dx = X0 * Tr = 3.83 * 0.328 = 1.26 m.
2. d = Vo*t + 0.5g*t^2 = 1 m.
0 + 4.9t^2 = 1,
t^2 = 0.204,
t = Tf = 0.452 s. = Fall time or time in flight.
a. Dx = Xo * Tf = 2 * 0.452 = 0.904 m.
b. Time of travel remains the same.
Dx = 4 * 0.452 = 1.81 m.
To solve these problems, we will use the equations of motion for projectiles, assuming there is no air resistance.
1. a) To find the maximum height reached by the ball from the floor, we can use the formula for vertical displacement:
\[y = y_0 + v_{0y}t - \frac{1}{2}gt^2\]
where:
- \(y\) is the vertical displacement
- \(y_0\) is the initial vertical position (1.5m)
- \(v_{0y}\) is the initial vertical velocity (calculated using \(v_{0y} = v_0 \sin(\theta)\))
- \(t\) is the time of flight (which we can solve for later using the formula \(t = \frac{2v_{0y}}{g}\))
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s^2)
First, let's calculate \(v_{0y}\):
\[v_{0y} = v_0 \sin(\theta) = 5 \sin(40^\circ)\]
Using a calculator, we find \(v_{0y} \approx 3.2\) m/s.
Now we can calculate the vertical displacement:
\[y = 1.5 + 3.2t - \frac{1}{2}(9.8)t^2\]
b) To find the horizontal distance required to reach the maximum height, we can use the formula for horizontal displacement:
\[x = v_{0x}t\]
where:
- \(x\) is the horizontal displacement
- \(v_{0x}\) is the initial horizontal velocity (calculated using \(v_{0x} = v_0 \cos(\theta)\))
- \(t\) is the time of flight
Let's calculate \(v_{0x}\):
\[v_{0x} = v_0 \cos(\theta) = 5 \cos(40^\circ)\]
Again, using a calculator, we find \(v_{0x} \approx 3.8\) m/s.
Now we can calculate the horizontal displacement:
\[x = 3.8t\]
c) To find the horizontal distance the ball traveled during this time, we can use the time of flight we found earlier (\(t = \frac{2v_{0y}}{g}\)):
\[d = 2xt\]
where:
- \(d\) is the horizontal distance traveled
- \(x\) is the horizontal displacement
- \(t\) is the time of flight
Plug in the values:
\[d = 2(3.8)t\]
2. a) To find the horizontal distance of the book from the table when it strikes the floor, we can use the equation for horizontal displacement:
\[x = v_{0x}t\]
where:
- \(x\) is the horizontal displacement
- \(v_{0x}\) is the initial horizontal velocity (which is equal to the initial speed of the book)
- \(t\) is the time of flight (which we can solve for later using the formula \(t = \sqrt{\frac{2h}{g}}\))
Plug in the values:
\[x = (2 \text{ m/s})t\]
b) To find the time of travel and the horizontal distance traveled if the initial speed is doubled, we have to consider the change in time and the change in horizontal velocity. We'll use the same equation for \(x\), but with the new values:
\[v_{0x} = v_0 \cos(\theta)\]
Doubling the initial speed means \(v_0\) is now \(4\) m/s.
After doubling the initial speed, we can find the new horizontal displacement \(x\) using the same equation:
\[x = (4 \text{ m/s})t'\]
To find the new time of travel \(t'\), we can use the formula we mentioned earlier:
\[t' = \sqrt{\frac{2h}{g}}\]
However, since only the initial speed is doubled, the height \(h\) remains the same (1m). So we can calculate \(t'\) using:
\[t' = \sqrt{\frac{2(1\text{ m})}{g}}\]
Now, with the new time of travel \(t'\), we can calculate the new horizontal distance:
\[x' = (4\text{ m/s})\cdot t'\]
To solve these problems, we can use basic equations of motion to analyze the projectile motion of the objects in question. Let's tackle each problem step by step.
1. Basketball Player:
a) To find the maximum height reached by the ball, we need to calculate the time it takes for the ball to reach its highest point. We can use the equation:
vf = vi + at
Since the ball reaches its highest point, its final vertical velocity (vf) will be 0 m/s. The initial vertical velocity (vi) is given as 5 m/s. Assuming the acceleration due to gravity is -9.8 m/s^2 (negative because it acts against the upward motion), we can rearrange the equation to solve for time (t):
0 = 5 - 9.8t
9.8t = 5
t ≈ 0.51 seconds
Now, we can use this time in another equation to find the maximum height (h) using the equation:
h = vi + 0.5at^2
h = 5(0.51) + 0.5(-9.8)(0.51)^2
h ≈ 1.26 meters
Therefore, the maximum height reached by the ball from the floor is approximately 1.26 meters.
b) To find the horizontal distance required to reach the maximum height, we can use the equation:
horizontal distance = horizontal velocity × time
The horizontal velocity remains constant throughout the projectile motion. Since the angle of release is given as 40 degrees above the horizontal, we can find the horizontal velocity using trigonometry:
horizontal velocity = initial speed × cos(angle of release)
horizontal velocity = 5 m/s × cos(40 degrees)
horizontal velocity ≈ 5 m/s × 0.766
horizontal velocity ≈ 3.83 m/s
Now, we can use this horizontal velocity and the time it takes to reach the maximum height (0.51 seconds) to find the horizontal distance:
horizontal distance = 3.83 m/s × 0.51 s
horizontal distance ≈ 1.95 meters
Therefore, the horizontal distance required to reach the maximum height is approximately 1.95 meters.
c) The horizontal distance the ball traveled during this time can be found by multiplying the horizontal velocity by the total time of flight. The total time of flight is twice the time it takes to reach the maximum height:
total time of flight = 2 × 0.51 seconds
total time of flight ≈ 1.02 seconds
horizontal distance = 3.83 m/s × 1.02 s
horizontal distance ≈ 3.91 meters
Therefore, the horizontal distance traveled by the ball during this time is approximately 3.91 meters.
2. Sliding Book:
a) To find the horizontal distance of the book from the table when it strikes the floor, we can use the equation of motion:
horizontal distance = initial horizontal velocity × time
The initial horizontal velocity remains constant throughout the motion. Since the book slides off the table horizontally, the initial horizontal velocity is given as 2 m/s.
The time of travel can be found using the equation:
vertical distance = initial vertical velocity × time + 0.5 × acceleration × time^2
Since the book falls freely under gravity, the initial vertical velocity is 0 m/s and the acceleration due to gravity is -9.8 m/s^2. The vertical distance fallen is 1 meter. Rearranging the equation, we get:
1 = 0 + 0.5 × (-9.8) × time^2
4.9 × time^2 = 1
time^2 = 1 / 4.9
time ≈ 0.45 seconds
Now, we can use this time in the horizontal distance equation:
horizontal distance = 2 m/s × 0.45 s
horizontal distance ≈ 0.9 meters
Therefore, the horizontal distance of the book from the table when it strikes the floor is approximately 0.9 meters.
b) If the initial speed is doubled, the new initial horizontal velocity would be 4 m/s.
To find the time of travel and the new horizontal distance, we can follow the same steps as before, but using the new initial horizontal velocity:
The time of travel:
1 = 0 + 0.5 × (-9.8) × time^2
4.9 × time^2 = 1
time^2 = 1 / 4.9
time ≈ 0.45 seconds
The horizontal distance:
horizontal distance = 4 m/s × 0.45 s
horizontal distance ≈ 1.8 meters
Therefore, if the initial speed is doubled, the time of travel remains the same (approximately 0.45 seconds), but the horizontal distance traveled is increased to approximately 1.8 meters.