A cannonball is located at the top of a 48.2 m building is shot horizontally with velocity of 112.1 m/s.Neglecting air resistance, The horizontal distance traveled by the ball? +y +x Answer:
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- Chinook salmon can cover more distance in less time by periodically making jumps out of the water. Suppose a salmon swimming in still water jumps out of the water with velocity 6.03 m/s at 40.6° above the horizontal, re-enters the water a distance L upstream, and then swims the same distance L underwater in a straight, horizontal line with velocity 2.92 m/s before jumping out again. What is the fish's average horizontal velocity (in m/s) between jumps? (Round your answer to at least 2 decimal places.) Consider the interval of time necessary to travel 2L. How is this reduced by the combination of jumping and swimming compared with just swimming at the constant speed of 92m/s? Express the reduction as a percentage. What If? Some salmon are able to jump a distance Lout of the water while only swimming a distance L/4 between jumps. By what percentage are these salmon faster than those requiring an underwater swim of distance L? (Assume the salmon jumps out of the water with velocity 03m/s…Dominador, who is standing on the edge of an 80-m tall cliff facing the sea, kicks a ball horizontally at a speed of 13.01 m/s. How far from the base of the cliff will the ball hit the sea water? Let be the initial vertical position of the ball, with positive position in the upward direction. The vertical displacement of the ball from the edge of the cliff to the seawater is: meters. The time it takes for the ball to reach the seawater is: t = seconds. The horizontal distance traveled by the ball is: meters.A ball is thrown upward at an angle of 28° to the horizontal with initial velocity ? = 62.0 m/s. Find: a.) The x and y components of ?. b.) The highest point of the ball’s path. c.) The horizontal distance that the ball travel. d.) The magnitude and direction of the ball velocity when t = 3.5 s.
- A marble is launched from the edge of a table at a angle of 45 ° up from horizontal and goes through the air until it hits the floor. True or False? For such a marble, |vx| (the magnitude of the x-component of the velocity) is decreasing the entire time the marble is airborne.A horizontal projectile is shot with an initial velocity of 17.9 m/s from a height of 2.59 m above the ground. Neglecting air-resistance, what is the magnitude of the final velocity (m/s)A ball is thrown with an initial velocity of 20 m/s at an angle of 60° above the horizontal. If we can neglect air resistance, what is the horizontal component of its instantaneous velocity at the exact top of its trajectory? 17 m/s 10 m/s 20 m/s zero
- A baseball pitcher throws a pitch with an initial velocity of 45 m/s (100 mph!), directed horizontally. You will be asked to find how far the ball drops vertically by the time it crosses the plate 18.0 m away. Which equation will you use to find the time it takes for the ball to travel the horizontal distance to the plate? Hint: Look at your given variables in the x-direction. vfx2=v0x2+2axΔx Δx =v0x t + ½ axt2 vfx=v0x+axt Δx=1/2(v0x+vfx)tA projectile, thrown horizontally, travels 100 m to hit back the ground. If its maximum height is S0 m then what is the initial velocity of the projectile? 3.19m/s 31.3m/s 10.2m/sIn the horizontal shooting test, if the height of the ball ejected from the assembly is y = 0.25m and the distance it travels in the horizontal is x = 0.65m, what is the launch speed (m / s) of the ball?
- A regulation volleyball court measures L = 18.0 meters in length and d = 2.43 meters in height. A volleyball player hits the ball from a height of h = 1.50 m just over the backline, and the initial velocity of the ball creates an angle of q = 30° with the ground. At what initial speed must the ball be struck such that it barely clears the net?Initially, a particle is moving at 5.40 m/s at an angle of 37.2 above the horizontal. Two seconds later, its velocity is 6.10 m/s at an angle of 55.0 below the horizontal. What was the particle's average acceleration during these 2.00 seconds in the x-direction and the y-directionA cannon ball is fired with an initial speed of 123 m/s at angle of 60 degrees from the horizontal. Express the initial velocity as a linear combination of its unit vector components. ( m/s) + ( m/s) At the maximum height, the speed of the cannon ball is v = ? m/s and the magnitude of its acceleration is a = ? m/s2. The time needed to reach maximum height is t = ? s. The maximum height reached by the cannon ball is H = ? m.