An archer aims directly at an apple attached to a tree branch a horizontal distance d away and a vertical height h up. At the exact moment the arrow is shot, the apple starts to fall from the tree. Show that (i.e. write down and solve equations for this situation that prove that) no matter the initial velocity of the arrow, the distance, or the height, it always hits the apple.
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- Water leaves a fireman’s hose (held near the ground) with an initial velocity v0 = 11.5 m/s at an angle θ = 30.5° above horizontal. Assume the water acts as a projectile that moves without air resistance. Use a Cartesian coordinate system with the origin at the hose nozzle position, as shown. 1). Using v0, θ, and g, write an expression for the time, tmax, the water travels to reach its maximum vertical height. 2)Consider a projectile being launched with an initial speed of 43 m/s at a variety of initial angles. Refer to the figure. What is the range, in meters, of the projectile if it is launched at an angle of θ1 = 79.7°? What is the range, in meters, of the projectile if it is launched at an angle of θ2 = 40.5°? What is the range, in meters, of the projectile if it is launched at an angle of θ3 = 90 − 79.7°, the complement of θ1?A projectile is fired with initial speed 150 m/s and an angle of elevation 45° from a position 10 m above ground level. Where does the projectile hit the ground and with what speed? Solution If we place the origin at ground level, then the initial position of the projectile is (0, 10) and so we need to adjust the parametric equations of the trajectory by adding 10 to the expression for y. With v0 = 150 m/s, ? = 45°, and g = 9.8 m/s2, we have x = 150 cos ? 4 t = y = 10 + 150 sin ? 4 t − 1 2 (9.8)t2 = . Impact occurs when y = 0, that is 4.9t2 − 75 2 t − 10 = 0. Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.) t = 75 2 + 11,250 + 196 9.8 ≈ Then x ≈ 75 2 (21.74) ≈ (rounded to the nearest whole number), so the projectile hits the ground about…
- How do you solve this? I've got 2/3 so farFrom the window of a building, a ball is tossed from a height yo above the ground with an initial velocity of 7.50 m/s and angle of 21.0° below the horizontal. It strikes the ground 6.00 s later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (Use the following as necessary: yo: Assume SI units. Do not substitute numerical values; use variables only.) X; = Y = (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. Vi, x Vi, y = m/s %3D %3D m/s (c) Find the equations for the x- and y-components of the position as functions of time. (Use the following as necessary: yo and t. Assume SI units.)You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity v at an angle 0 with respect to the horizontal. Let the building be 54.0 m tall, the initial horizontal velocity be 9.10 m/s, and the initial vertical velocity be 10.5 m/s. Choose your coordinates such that the positive y-axis is upward, and the x-axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, find the ball's maximum height above the ground and the time it takes to reach the maximum height. maximum height above ground time to reach maximum height (b) Repeat your calculations choosing the origin at the base of the building. maximum height above ground time to reach maximum height
- Use a graphing utility to obtain the path of a projectile launched from the ground (h = 0) at the specified values of θ = 35°, v0 = 300 feet per second. In the exercise, use the graph to determine the maximum height and the time at which the projectile reaches its maximum height. Also use the graph to determine the range of the projectile and the time it hits the ground. Round all answers to the nearest tenth.This problem will involve deriving a formula or two for a projectile launched from one height and angle and landing at a different height on Earth. Begin with a projectile launched at angle 0 above horizontal from a height y₁ with initial velocity Vo. The projectile lands at a point with height y₂. These are the given quantities: vo, 0, y₁, y2 and g. Construct formulae for each of the following, as. a function of given quantities the horizontal distance traveled. the maximum height reached. the time taken. the angle of impact. (find the final velocity components first).Projectile #1 is launched at t%=DOs at position (0,0) with an initial x-velocity of 5m/s and y-velocity of 10m/s. Projectile #2 is launched 1s after projectile #1 from an initlal position of (1,1) with an x-velocity of 10m/s. Find the following: The position that the 2 projectiles collide at, the time it takes to collide, and the Initial velocity of projectile #2 Including the launch angle.
- Question 2. A projectile is fired from the ground with an initial speed v at an angle 0 above the horizontal. (a) At t = 0, what is the velocity vector? (b) At what time t' does the projectile reach its maximum height? (c) What is the velocity vector at the top of the trajectory? (d) What is the horizontal distance traveled by the projectile? (e) At what angle should one launch the projectile to maximize horizontal distance traveled?Projectile A is launched at an angle of 20 degrees to the horizontal with initial speed u. Projectile B is launched at an angle of 70 degrees to the horizontal, also with an initial speed u.(a) Compare the times of flight of the two projectiles.(b) Show that the ranges of both projectiles are equal.A river of width' a'with straight parallel banks flows due north with speed u. The points O and A are on opposite banks and A is due east of O. Coordinate axes Ox and Oy are taken in the east and north directions respectively. A boat, whose speed is v relative to water, starts from O and crosses the river. If the boat is steered due east and u varies with x as : u = x(a – x)- Find : absolute velocity of boatman when he reaches the opposite bank