A motorcycle daredevil is attempting to jump across as many buses as possible (see the drawing). The takeoff ramp makes an angle è = 18.0° above the horizontal, and the landing ramp is identical to the takeoff ramp. The buses are parked side by side, and each bus is 2.74 m wide. The cyclist leaves the ramp with a speed of v = 30.0 m/s. What is the maximum number of buses over which the cyclist can jump?
sick
To determine the maximum number of buses the cyclist can jump over, we need to calculate the distance covered by the motorcycle in the horizontal direction during the jump.
First, let's find the horizontal component of the motorcycle's velocity, denoted as Vx. To do this, we'll use the formula:
Vx = v * cos(è)
Where v is the initial speed of 30.0 m/s and è is the takeoff angle of 18.0°.
Vx = 30.0 m/s * cos(18.0°) = 28.07 m/s
Next, we'll calculate the time it takes for the motorcycle to complete the jump. To do this, we'll find the time it takes for the motorcycle to land. Since the landing ramp is identical to the takeoff ramp, the landing angle is also 18.0°.
Using the kinematic equation:
Δy = Vyt + 1/2 * a * t^2
Where Δy is the vertical distance, Vyt is the vertical component of the velocity at the time of landing, a is acceleration (assumed as -9.8 m/s^2 due to gravity), and t is time.
Since the motorcycle lands at the same vertical position it takes off (assuming no air resistance), Δy = 0. The vertical velocity at the time of landing (Vyt) can be determined using the equation:
Vyt = v * sin(è)
Vyt = 30.0 m/s * sin(18.0°) = 9.81 m/s
We can now solve for t:
0 = 9.81 m/s * t + 1/2 * (-9.8 m/s^2) * t^2
Rearranging the equation, we get:
1/2 * -9.8 m/s^2 * t^2 + 9.81 m/s * t = 0
Since the coefficient of t^2 is negative, the solutions for t will have opposite signs. However, we're interested in the positive time value, so we have:
t = 0 s or t = 2 * (9.81 m/s) / 9.8 m/s^2 = 2.00 s
The total time for the jump is then:
t_total = 2 * t = 2 * 2.00 s = 4.00 s
Finally, we can calculate the horizontal distance covered by the motorcycle, denoted as d:
d = Vx * t_total = 28.07 m/s * 4.00 s = 112.28 m
Each bus is 2.74 m wide, so the maximum number of buses the cyclist can jump over is given by:
Number of buses = d / bus width
Number of buses = 112.28 m / 2.74 m = 41 buses (rounded down to the nearest whole number since the cyclist cannot land partially on a bus)
Therefore, the maximum number of buses the cyclist can jump over is 41.
To determine the maximum number of buses the cyclist can jump over, we need to find the distance the cyclist can cover during the jump.
Let's analyze the situation:
1. The takeoff angle (θ) is given as 18.0° above the horizontal.
2. The landing ramp is identical to the takeoff ramp, so the landing angle is also 18.0° above the horizontal.
3. The width of each bus (w) is given as 2.74 m.
4. The speed of the cyclist (v) is given as 30.0 m/s.
To find the maximum number of buses the cyclist can jump, we first need to find the horizontal distance covered during the jump.
The horizontal distance (d) is given by:
d = v * t
where t is the time of flight.
To find t, we need to determine the time it takes for the cyclist to reach the maximum height. At the maximum height, the vertical component of the cyclist's velocity becomes zero.
The vertical component of the initial velocity is given by:
v_vertical = v * sin(θ)
At the maximum height, the vertical component of the velocity becomes zero:
0 = v * sin(θ) - g * t_max
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t_max is the time to reach the maximum height.
Solving for t_max, we get:
t_max = (v * sin(θ)) / g
Using this value of t_max, we can find the total time of flight (t) using:
t = 2 * t_max
The horizontal distance covered during the jump (d) is then found using:
d = v * t
Now, we can calculate the maximum number of buses the cyclist can jump by dividing the horizontal distance (d) by the width of each bus (w):
max_buses = d / w
Let's plug in the values and calculate the result:
θ = 18.0°
w = 2.74 m
v = 30.0 m/s
g = 9.8 m/s^2
sin(θ) = sin(18.0°) ≈ 0.309
t_max = (30.0 * 0.309) / 9.8 ≈ 0.948 s
t = 2 * 0.948 ≈ 1.896 s
d = 30.0 * 1.896 ≈ 56.88 m
max_buses = 56.88 / 2.74 ≈ 20.8
Therefore, the maximum number of buses the cyclist can jump is approximately 20 buses. However, since we cannot have a fraction of a bus, the cyclist can jump over a maximum of 20 buses.
horizontal speed = u = constant = 30 cos 18 so find u
d = u T
so we need T (total time aloft) to know the horizontal distance, d in the air.
so how long in the air?
initial speed up = Vi = 30 sin 18
so find Vi
v = Vi - 9.8 t
at top v = 0 and t = T/2
so
T/2 = Vi/9.8
T = Vi /4.9
put that T back in d = u T to find how far it goes in the air