You want to jump from an 8.2 m high building to a 10 m high building. The gap between the buildings is 2.4 m, and you can run fast enough so that your speed as you leave the edge of
the lower building is 7.0 m/s. What range of launch angles will allow you to reach the top of the
second building?
I don't even know where to start with this. I know that the horizontal velocity would be 7cos(theta) and the vertical velocity would be 7sin(theta)-9.8t. Which then allows me to find that the horizontal position is 7cos(theta)t, and the vertical position is 7sin(theta)t-4.9t^2. What I don't know is how to find if what angle range leaves the horizontal position greater than 2.4 and the vertical position is greater than 1.8. Any help is great, thank you.
Lol, do you go to ISU or something?
Range = Vo^2*sin(2A)/g.
2.4 = 7^2*sin(2A)/9.8.
A = 14.7o. = Minimum launching angle.
Max. range occurs at 45o:
Range = 7^2*sin(2*45)/9.8 = 5 m., max.
Range of launch angles = 14.7 t0 45o.
To find the range of launch angles that will allow you to reach the top of the second building, you need to solve two conditions simultaneously: the horizontal position being greater than 2.4 meters, and the vertical position being greater than 1.8 meters.
First, let's analyze the horizontal position. The horizontal position is given by the equation:
x = v₀ₓ * t
where v₀ₓ is the horizontal component of the initial velocity, and t is the time of flight. In this case, v₀ₓ is equal to 7.0 m/s since the initial velocity is given as 7.0 m/s and there is no horizontal acceleration. The time of flight can be determined by dividing the horizontal distance (2.4 meters) by the initial horizontal velocity:
t = x / v₀ₓ = 2.4 m / 7.0 m/s
Next, let's analyze the vertical position. The vertical position is given by the equation:
y = v₀ᵧ * t - (1/2) * g * t²
where v₀ᵧ is the vertical component of the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time of flight. In this case, v₀ᵧ can be determined by multiplying the initial vertical velocity (which is 0 m/s since you are starting from the ground) by the sine of the launch angle (θ):
v₀ᵧ = 0 m/s * sin(θ) = 0 m/s
Substituting these values into the equation, we get:
y = - (1/2) * g * t² = - (1/2) * 9.8 m/s² * (2.4 m / 7.0 m/s)²
Now, rearrange the equation to solve for θ:
θ = arcsin[(2 * y * g * t) / (v₀ᵧ * t)²]
θ = arcsin[(2 * 1.8 m * 9.8 m/s² * (2.4 m / 7.0 m/s)) / (0 m/s * (2.4 m / 7.0 m/s))²]
Evaluating this expression will give you the launch angle (θ). Keep in mind that the sine function has a range of -1 to 1, so the expression inside the arcsin should be within this range. If the expression exceeds 1, it means that it is not physically possible to reach the top of the second building given the initial conditions.
Lastly, you may want to consider the fact that the result of arcsin will give you an angle in radians. If you need the angle in degrees, you can convert it using the formula:
θ (degrees) = θ (radians) * (180/π)
Now you can use this approach to find the range of launch angles that will allow you to reach the top of the second building.