Consider an object sends into projectile from ground level with initial velocity of voand angle from horizontal axis. a) Find an equation for time to reach the maximum height of the projectile? b) Find an equation for maximum height of the projectile? c) Find an equation for the range of the projectile?
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- Problem For a projectile lunched with an initial velocity of vo at an angle of e (between O and 90), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of e gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x= vocos(e) Vinitial-y = vosin(e) 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 Thus, the time to reach the maximum height is Imax-height We will use this time to the equation yfinal - Yinitial = Vinitialyt+ (1/2)a,2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-y + (1/2lay? substituting the Vinitial-y expression above, results to the following…Problem For a projectile lunched with an initial velocity of vo at an angle of 0 (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of e gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = Vocos(0) Vinitial-y = Vosin(e) 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 %3D 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)ayt 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)ay? substituting, the vinitial-y expression above, results to…a projectile is launched at an angle of 35 degrees with an initial velocity of 780 ft/s from the ground. a) what is the velocity at t=3s b) what is the total time of flight? c) what is the horizontal distance from the origin when projectile is at it's maximum height?
- The exit velocity of a baseball (its velocity as it leaves the bat) is 128 feet per second, in the direction of 30 above horizontal and directly towards the monster 30 fcot tall center field wall. If the ball was hit 4 feet above ground level, find a) Parametric equations that model the trajectory of the baseball, b) The maximum height of the baseball's fiight, c) If the center field fence is 400 feet from home plate is this a home run? d) What is the impact velocity of the ball? =) Sketch a graph representing the trajectory of the ball.Using the table above, Plot the: a. Maximum range of the projectile as a function of the projection angle and b.maximum vertical displacement of the projectile as a function of the projection angleConsider an object sends into projectile from ground level with initial velocity of v 0 and θ angle from horizontal axis. a) Find and equation for time to reach the maximum height of the projectile?
- 1) A dog. initially at position 2.8 m [W] of its owner, runs to retrieve a stick that is 12.6 m [E] of its owne. a) Draw an illustration showing the position and displacement vectors in this situation. b) Determine the displacement the dog needs to reach the stick. 2) The table below gives the position-time data of a ball that has left a bowler's hand and is rolling forward. Detemine the displacement between the following times: (Please refer to the photo) 3) A motorcycle took only 6.0 seconds to go from rest to 78 m/s (2.8 x 102 km/hr), earning the world record for motorcycle acceleration. Calculate the magnitude of the record average acceleration in m/s2Problem 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 2.54m tall tennis player is 11.72m away from the net of 0.90m inheight. If the ball clears a height of 0.15m above the net and the player strikes theball horizontally from his initial position, determine: a. velovity of a ball b. the distance measured from the net where the ball hits the other side of the courtA ball kicked from ground level at an initial velocity of 60 m/s and an angle 0 with ground reaches a horizontal distance of 200 meters. a) What is the size of angle 0? b) What is time of flight of the ball?A projectile is lunched at velocity (600 ) with an angle (60°). Find:- sec. a) Horizontal range. The velocity and height of projectile after (30 sec). From lunching. Maximum height reaches of projectile. b) c)