A daredevil performs a stunt in which he jumps with his motorcycle from a horizontal cliff 10m high. What must the initial velocity of the motorcycle be, in order for the stuntman to barely clear a 10m wide water canal at the base of the cliff?
distance across=vhorizongal*time
now, in vertical, d=1/2 g t^2 OR
t=sqrt(20/9.8)
put that time in the first equatioan, solve for vhorizontal
I got the wrong answer. My answer key says the answer is 7m/s.
t = sqrt(20/9.8) = 1.428
d = 1/2(9.8)(1.428)^2 = 10
d*t = 14.28
What am I doing wrong?
h = 0.5g*t^2 = 10 m.
h = 4.9*t^2 = 10
t^2 = 2.04
Tf = 1.43 s
Xo*1.43 = 10 m.
Xo = 7.0 m/s
To find the initial velocity of the motorcycle, we can use the principles of projectile motion. Let's break down the problem into two parts: vertical motion and horizontal motion.
First, let's consider the vertical motion. The stuntman must jump from a 10m high cliff and barely clear a 10m wide water canal. Ignoring air resistance, we can use the equation for vertical displacement in projectile motion:
Δy = V₀t + (1/2)gt²
Where:
Δy is the vertical displacement (10m)
V₀ is the initial vertical velocity (which we need to find)
t is the time of flight (which will be the same for both horizontal and vertical motions)
g is the acceleration due to gravity (approximately 9.8 m/s²)
Since the stuntman must barely clear the 10m wide canal, we want the vertical displacement at the time it reaches the canal to be 0. Therefore:
0 = V₀t - (1/2)gt²
We can rearrange this equation:
t = (2V₀/g)
Now, let's consider the horizontal motion. The horizontal displacement is given as 10m, which is equal to the width of the canal. The horizontal distance can be calculated using the equation:
Δx = V₀ₓt
Where:
Δx is the horizontal displacement (10m)
V₀ₓ is the initial horizontal velocity (which we need to find)
t is the time of flight (which we already determined as (2V₀/g))
Substituting the values:
10 = V₀ₓ(2V₀/g)
Simplifying the equation:
V₀ₓ = 5g/V₀
Lastly, we need to find the total initial velocity V₀. Since the motion is a combined motion of vertical and horizontal, we can use the Pythagorean theorem:
V₀² = V₀ₓ² + V₀y²
Since we know that the vertical and horizontal components of the velocity will be equal (since the motorcycle lands at the same height as it takes off), we can simplify the equation:
V₀² = (5g/V₀)² + V₀²
Simplifying further:
V₀² = 25g²/V₀² + V₀²
Combining like terms:
2V₀² = 25g²/V₀²
Finally, solving for V₀:
2V₀⁴ = 25g²
V₀⁴ = 12.5g²
Taking the fourth root on both sides:
V₀ = √(√(12.5g²))
From here, you can substitute the value of acceleration due to gravity (g) as 9.8 m/s² and calculate the initial velocity (V₀).