You must have seen actors in Hindi films jumping over huge gaps on horse backs and motor cycles. In this problem consider a daredevil motor cycle rider trying to cross a gap at a velocity of 100 km h–1. (Fig. 4.13). Let the angle of incline on either side be 450. Calculate the widest gap he can cro
Ah, the daredevil motorcyclists of Hindi films, always defying gravity! Now, let's calculate the widest gap this brave rider can cross.
To do that, we need to consider the physics involved. We have the velocity of 100 km/h and an incline angle of 45 degrees on either side.
First, let's convert the velocity from km/h to m/s. We multiply 100 km/h by 1000/3600 to get approximately 27.78 m/s.
Now, let's break down the motion into two components: horizontal and vertical.
In the horizontal direction, the velocity remains constant. So, we can use the formula v = d/t, where v is the horizontal velocity, d is the distance, and t is the time. Rearranging the formula, we get d = v * t.
In the vertical direction, we have an initial velocity of zero and the acceleration due to gravity acting downward. We'll use the formula h = (1/2) * g * t^2, where h is the vertical height, g is the acceleration due to gravity, and t is the time.
Since we have symmetric inclines, we can assume the time taken to climb and descend each incline is the same. Let's call this time as 't'. Hence, the total time for the complete jump is 2 * t.
Now, let's plug in the numbers. The horizontal distance 'd' is the sum of the two gaps, so d = d1 + d2.
For each gap, let's consider only the time taken to climb one incline and multiply it by 2 to get the total time.
Using the formula v = d/t, we can rewrite it as t = d/v.
For the first gap (d1), t1 = d1/v, and for the second gap (d2), t2 = d2/v.
Now, let's calculate the maximum vertical height 'h' using the formula h = (1/2) * g * t^2.
Since we have the same time for each gap, the height h is the same for both inclines.
Finally, just add the two heights to get the total width of the gap he can cross, i.e., widest gap = 2h.
Phew! That was quite a calculation, wasn't it? But remember, safety first! Leave the acrobatics to the experts on the big screen.
I dont get your angle of incline b eing 450?
lets call it theta.
initial vertical speed=100km/hr*1hr/3600sec*1000m/km(*sinTheta=27.7(sinTheta) m/s
initial horizontal speed=27.7(cosTheta)
At the widest gap, he will land at the edge (barely), so jump distance=horizontal speed*timeinair.
timeinair: Hf=hi+vi'cosTheta*time-1/2 g time^2
or time(vi-4.9t)=0 or timeinair= 27.7*sinTheta*/4.9 check that.
widest gap= 27.7*cosTheta*27.7*sinTheta/4.9
if you wish, you can convert sin*cos to another angle using the addition formula, but i see no advantage to doing so.. Note the max width will occur at 45 deg
To calculate the widest gap the motorcyclist can cross, we can use the concept of horizontal projectile motion.
Given:
Velocity of the motorcyclist, v = 100 km/h
Angle of incline on either side, θ = 45°
To solve this problem, we can break the motion into horizontal and vertical components.
1. Horizontal Component:
The horizontal component of velocity remains constant during the entire motion. We can find the horizontal component of velocity, vx, using trigonometry:
vx = v * cos(θ)
Here, v is the velocity in km/h and θ is the angle of incline. Let's substitute the given values and calculate vx.
vx = 100 km/h * cos(45°)
= 100 km/h * 0.7071
≈ 70.71 km/h
2. Vertical Component:
The vertical component of velocity changes due to the force of gravity. We can find the vertical component of velocity, vy, using trigonometry:
vy = v * sin(θ)
Here, v is the velocity in km/h and θ is the angle of incline. Let's substitute the given values and calculate vy.
vy = 100 km/h * sin(45°)
= 100 km/h * 0.7071
≈ 70.71 km/h
3. Maximum Gap:
To determine the maximum gap the motorcyclist can cross, we need to find the time it takes for the motorcyclist to reach the highest point of the trajectory. At the highest point, the vertical component of velocity becomes zero.
Using the vertically upward motion equation:
0 = vy - g * t
Here, vy is the initial vertical velocity, g is the acceleration due to gravity, and t is the time taken to reach the highest point.
Substituting the values:
0 = 70.71 km/h - (9.8 m/s^2) * t
Solving for t:
t = (70.71 km/h) / (9.8 m/s^2)
= (70.71 km/h) / (9.8 * 1000 m/s^2)
≈ 0.007 m/s
Now, to find the maximum gap, we can use the horizontal motion equation:
s = vx * t
Here, s is the maximum gap, vx is the horizontal velocity, and t is the time taken to reach the highest point.
Substituting the values:
s = 70.71 km/h * 0.007 s
≈ 0.494 km
Therefore, the widest gap the motorcyclist can cross is approximately 0.494 km.
To calculate the widest gap the motor cycle rider can cross, we can use the concept of projectile motion. The key here is to find the range of the projectile, which is the horizontal distance traveled by the rider.
First, let's convert the velocity from km/h to m/s. Since 1 km = 1000 m and 1 h = 3600 s, we have:
100 km/h = (100 * 1000) m / (3600 s) = 250/9 m/s (approximately)
Now, let's consider the horizontal and vertical motions separately.
Horizontal Motion:
Since the rider is traveling at a constant horizontal velocity, there is no acceleration in the horizontal direction. The time taken to cross the gap (t) is simply the distance divided by the horizontal velocity:
t = Distance / Horizontal Velocity
Vertical Motion:
The rider is moving up the incline, which means his initial vertical velocity (Vy) is given by:
Vy = Velocity * sin(angle of incline)
The time taken to reach the highest point (t1) can be calculated using the equation:
0 = Vy - g * t1
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
The time taken to descend down the incline (t2) can be calculated using the equation:
0 = -Vy - g * t2
Now, let's calculate the time taken for both vertical motions (t1 and t2) and the total time taken (t) to cross the gap.
t1 = Vy / g
t2 = -Vy / g
t = t1 + t2
Next, we can calculate the range (R) using the horizontal velocity and the total time taken:
R = Horizontal Velocity * t
Finally, we have the widest gap the rider can cross, which is equal to the range (R) of the projectile.
Therefore, to calculate the widest gap the motor cycle rider can cross, substitute the given values (velocity, angle of incline) into the equations above and calculate accordingly.