A secret agent skis off a slope inclined at θ = 31.5 degrees below horizontal at a speed of v0 = 14.7 m/s. He must clear a gorge, and the slope on the other side of the gorge is h = 18.4 m below the edge of the upper slope. What is the maximum width, w, of the gorge (in meters) so that the agent clears it?

I need to solve v(sub y)=v(sub 0)sin θ(sub 0)-gt to find the time that it takes the projectile to reach maximum height (so I believe I'm solving for t?). I have previously found v(sub 0y) in part (a), but I'm unsure what v(sub 0) and v(sub y) are. How do I find those variables?

A sailboat sails from Marblehead Harbor directly east for 45 nautical miles, then 60 degree south of east for 20 nautical miles, returns to an easterly heading for 30 nautical miles, and sails 30 degreeeast of north for 10 nautical miles, then west for 62 nautical miles. At that time the boat becomes becalmed and the auxiliary engine fails to start. The crew decides to notify the Coast Guard of their position. What is their position?

A small ship capable of making a speed of 6.5 knots through still water maintains a heading due east while being set to the south by an ocean current. The actual course of the boat is from A to B, a distance of 8.3 nautical miles that requires exactly 1.6 hours. Determine the speed v. of the current and its direction 0 measured clockwise from the north. N(0°) A -E(90°) 11 Answers: Vc = i knots i

Problem 3: An arrow is shot from a height of 1.8 m toward a cliff of height H. It is shot with a velocity of 35 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.7 s later. What is the height H of the cliff in m? H = cos() asin() sin() tan() 7 8 9 HOME cotan() acos() 5 atan() acotan() sinh() 1 2 3 cosh() tanh() cotanh() + END ODegrees O Radians vol BACKSPACE DEL CLEAR Submit Hint Feedback I give up! What is the maximum height reached by the arrow along its trajectory in meters? What is the arrow's speed just before hitting the cliff in m/s?

The citizens of Paris were terrified during World War I when they were suddenly bombarded with shells fired from a long-range gun known as Big Bertha. The barrel of the gun was 36.6 m long, and it had a muzzle speed of 2.20 km/s. When the gun’s angle of elevation was set to 55.0°, what would be the range? For the purposes of solving this problem, ignore air resistance. (The actual range at this elevation was 121 km; air resistance cannot be ignored for the high muzzle speed of the shells.) km

The drawing shows an exaggerated view of a rifle that has been 'sighted in' for a 91.4-meter target. If the muzzle speed of the bullet is Vo = 309 m/s, there are the two possible angles 0₁ and 02 between the rifle barrel and the horizontal such that the bullet will hit the target. One of these angles is so large that it is never used in target shooting. Give your answers as (a) the smaller angle and (b) the larger angle. (Hint: The following trigonometric identity may be useful: 2 sine cose = sin 20.) (a) Number (b) Number Units Units 91.4 m O

An object has an initial velocity of 29.0 m/s at 95.0° and an acceleration of 1.90 m/s2 at 200.0°. Assume that all angles are measured with respect to the positive x-axis. (a) Write the initial velocity vector and the acceleration vector in unit vector notation. (b) If the object maintains this acceleration for 12.0 seconds, determine the average velocity vector over the time interval. Express your answer in your unit vector notation.

The banded archerfish is a species of fish that lives in mangrove estuaries in Asia and Oceania. It has a unique and highly effective hunting strategy: it shoots an incredibly precise stream of water out of its mouth at almost ten meters per second, knocking insects and other small animals into the water from nearby branches! Pom Fbug (t) Ө = Our hero, a hungry archerfish, has spotted a big, delicious bug sitting on a branch a height ħ above the surface of the water. The archerfish can shoot its water jet at a speed of vo. The archerfish wants to knock the bug sideways off of the branch, so it decides to shoot so that its water jet is moving horizontally at the moment when it strikes the bug. The final goal of this problem is to find the horizontal distance, d, from the branch, and the angle above horizontal, 0, at which archerfish should shoot. d (a) What are the position and velocity of the water droplet as a function of time and the position and velocity of the bug as a function of…

-b+√b²-4ac 2a x = x₂ + V₁xt + ²axt², V₁₁x = V₁x + axt, (v₁₁x)² = (v₁x)² + 2axx, t == у = V₁ + V₁¸yt + ² α¸t², Vf₁y = V₁y+āÿt, (Vƒ‚y)² = (v₁y)² + 2a,ÿ