The ball is kicked from a ground level with an initial velocity of 10 m/s at an angle of 30 degrees. Determine: a.) The Range; b.) Speed of the ball when it strikes the ground.
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- A projectile was launched vertically upward with an initial speed of 29.2 m/s from the ground. a) Find the time it takes to get to the maximum height. b) Find the maximum height of the projectile. c) Find its position 2 seconds later. 1. d) When is the speed of the projectile V=-30m/s1. A golf ball is hit from 4.3 m above a golfing fairway with an initial velocity of 30.0 m/s at an angle of 35° above the horizontal. a) Determine the time of flight for the ball. b) Determine the range for the golf ball. c) Determine the velocity for the golf ball the instant before the ball impacts the ground.Problem: A golfer strikes a golf ball with an initial velocity of 45 m/s at an angle of 40° to the horizon. a) Calculate the initial x velocity. b) Calculate the initial y velocity. c) What is the balls velocity just before it strikes the ground? d) What is the y velocity at the apex of the ball’s motion? e) Calculate the total time the ball will be in the air. f) Calculate the maximum range of the ball (how far does it go?) g) Calculate the maximum height reached by the ball? Show all work!
- A ball is kicked with a velocity of 30 m/s at an angle of 40 degrees with the ground. Determine the range and maximum height of the ball.The world long jump record is 8.95m. Treated as a projectile, what is the maximum range obtained by a person if he has a take off speed of 9.5m/s? State your assumptionsFrom the edge of the roof of a 40 m high building an object is thrown with a speed v0 = 70 m / s and a throw angle with respect to the horizontal of 55 °. Determine: a) The horizontal and vertical components of the initial velocity. b) The distance from the base of the building at which the object will fall to the ground c) The time the object is in the air from when it is thrown until it reaches the ground. d) The maximum height that the object will reach above the ground. e) The velocity v of the object at the moment it hits the ground. f) The angle a with respect to the vertical with which the object reaches the ground.
- A golfer hits a golfball off a cliff from 8.5 metres above flat ground. The golfball is hit with an initial velocity of 43m/s [33 degrees above the horizontal]. a) Calculate the time of flight of the golfball. b) Calculate the horizontal range of the golfball. c) Calculate the maximum height of the golfball above the ground. d) Calculate the velocity of the golfball at its maximum height.1 ) A ball kicked from ground level at an initial velocity of 69 m/s and an angle θ with ground reaches a horizontal distance of 150 meters. a) What is the size of angle θ? b) What is time of flight of the ball?Problem For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: vfinal-y = vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is vfinal-y = ? Then, ? = vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: ? = ? - ?t Thus, the time to reach the maximum height is tmax-height = ?/? We will use this time to the equation yfinal - yinitial = vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = vinitial-yt + (1/2)ayt2 substituting, the vinitial-y…
- Problem For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) a) Let us first find the time it takes for the projectile to reach the maximum height.Using: vfinal-y = vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height isvfinal-y =___________ Then, ___________= vinitial-y + aytSubstituting the expression of vinitial-y and ay = -g, results to the following: __________=__________-__________t Thus, the time to reach the maximum height istmax-height = __________ /__________ We will use this time to the equation yfinal - yinitial = vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax =…A ball is launched at a velocity of 20 m/s in a direction making an angle of 25 upward with the horizontal. a) What is the maximum height reached by the ball? b) What is the total flight time (between launch and touching the ground) of the ball? c) At what horizontal distance from the point of release will it strike the ground? d) What is the magnitude of the velocity of the ball just before it hits the ground?A projectile is fired from point A at the edge of a cliff, whose initial velocity components are v0x = 60.0m / s and v0y = 175m / s. The projectile rises and then falls into the sea at point P. The projectile flight time is 40.0s and it does not experience appreciable air resistance in flight. a) Make a diagram of the problem, b) Determine how far it falls, c) what is the height of the cliff? and d) at what angle was it shot?
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