A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the offi?
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials' boat pulls up next to the spy's boat, both reach the edge of a 4.0 m waterfall. If the spy's speed is 13 m/s and the officials' speed is 23 m/s, how far apart will the two vessels be when they land below the waterfall?
A spy in a speed boat is being chased down a river by government officials in a faster craft.?
First, we need to find the time of flight for the boats. We use the equation for the Time of Flight for a horizontally thrown projectile: T = √(2h/g) Note that this is the reduced form of the larger equation as here the launch angle(θ) is zero (So the v sinθ and the (v sinθ)^2 terms vanish) Putting Platform height h = 6.7 m (Height of waterfall), we get time of Flight T = 1.169336 sec , which is gonna be the same for both boats. Now we find out where both the boats landed from the equation for Range of Projectile: R = vT cos(θ) For the spy's boat, putting Velocity v = 19 m/s, launch angle θ = 0 and Time of flight T as found earlier = 1.169336 sec, we get spy's position from the waterfall: R1 = 19 * 1.169336 * 1 (As cos 0 = 1) ∴ R1 = 22.217384 m Similarly, for finding where the officials' boat lands, put velocity v = 23 m/s keeping other terms same to get: R2 = 26.894728 m ∴ The distance between them when they fall is: R2 - R1 = 26.894728 - 22.217384 = 4.677343 m
A spy in a speed boat is being chased down a river by government officials in a faster craft.?
Just as the officials’ boat pulls up next to the spy’s boat, both boats reach the edge of a 5.3 m waterfall. The spy’s speed is 14 m/s and the officials’ speed is 26 m/s. How far apart will the two vessels be when they land below the waterfall? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m
A spy in a speed boat is being chased down a river by government officials in a faster craft. just as the offi?
The procedure here is to solve for time and then to use time to solve for range. Then compare the ranges. First, solve for time using y = yo + (voy)t - (0.5)gt^2. We know: y = 0m yo = 5.0 m voy = 0 m/s (because the projectile is being launched horizontally, it has no initial vertical velocity) t = ? g = 9.8 m/s^2 0 = 5 - (0.5)(9.8)t^2 t = 1.01 s Now, we know that the slower boat travels at a rate of 15 m/s for 1.01 s. So: R = D / T 15 = D / 1.01 D = 15.15 m The faster boat travels at a rate of 26 m/s for 1.01 s. So: R = D / T 26 = D / 1.01 D = 26.26 m 26.26 - 15.15 = 11.11 m EDIT: haha Hannah, you were right with the 11
A spy in a speed boat is being chased down a river?
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials’ boat pulls up next to the spy’s boat, both boats reach the edge of a 5.0 m waterfall. The spy’s speed is 13 m/s and the officials’ speed is 28 m/s. How far apart will the two vessels be when they land below the waterfall? The acceleration of gravity is 9.81 m/s2 . Answer in units of m = ?
Physics help?! A spy in a speed boat is being chased down a river by government officials in a faster craft. J?
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials' boat pulls up next to the spy's boat, both reach the edge of a 7.2 m waterfall. If the spy's speed is 15 m/s and the officials' speed is 27 m/s, how far apart will the two vessels be when they land below the waterfall?
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the offi?
First you need to figure out how long each boat is in the air after they go over the fall. d = vt + (1/2)at^2 d in this case is the VERTICAL distance, and v is the DOWNWARD velocity. The vertical distance is 5.2m, the initial downward velocity is 0, and the acceleration is 9.81. We can now solve for t: 5.2 = 0t + (1/2)(9.81)t^2 5.2 = 4.905*t^2 1.06 = t^2 1.03 = t (I'm rounding everything) Now that you know how long the boats are in the air (t = 1.03), just figure out how far each boat goes HORIZONTALLY in that time. spy boat = 14 m/s for 1.03s = 14.42m gov't boat = 28 m/s for 1.03s = 28.84m And the difference between those two numbers is 14.42m. So the boats will be 14.42m apart.
A spy in a speed boat is being chased down a river...WEBASSIGN PHYSICS QUESTION!!?
for the spy: v_initial = 20m/s Acceleration along x-axis is 0. Acceleration along y-axis is 9.8 (g). in y-direction: d = vi t + 1/2gt^2. vi along y-axis is zero so t = 1.25s Distance along x direction would be vi*t = 25m Same logic for official's boat speed, so the result would be 28.8 m. Difference is 3.76m.