A ball is thrown horizontally at an angle with an initial velocity of initial vertical velocity of 29.4 m/s. The trajectory diagram shows the position of the ball after each consecutive second. Apply your understanding of projectiles by filling in the blanks. Vy 11 Note: This vector 10 ら- diagram is depicting velocity vectors, not force or acceleration 3 6. Vy 6. vectors 2 13 -29.4/s 8 V- 10 m/s Vy 14

A ball is thrown horizontally at an angle with an initial velocity of initial vertical velocity of 29.4 m/s. The trajectory diagram shows the position of the ball after each consecutive second. Apply your understanding of projectiles by filling in the blanks. Vy 11 Note: This vector 10 ら- diagram is depicting velocity vectors, not force or acceleration 3 6. Vy 6. vectors 2 13 -29.4/s 8 V- 10 m/s Vy 14

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Description: Put the answers on every blank in each number pls. I don't want to waste my money here. A ball is thrown horizontally at an angle with an initial velocity of 10.0 m/s and aninitial vertical velocity of 29.4 m/s. The trajectory diagram shows the posit

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter4: Motion In Two And Three Dimensions

Section: Chapter Questions

Problem 59P: Aaron Rodgers throws a football at 20.0 m/s to his wide receiver, who runs straight down the field...

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Situation t1- A projectile is fired with an imitial velocity 4 320% at an argle 4 sd with the horizontal. Reg'd. a-Position y A ard h, at the 'erd of the projectile 6-Velscity ard direction 4 the tion af the erd f 105. Vo-320m/s
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. Prove that the trajectory of a projectile is parabolic, havingthe form y = ax + bx2. To obtain this expression, solve theequation x = v0xt for t and substitute it into the expressionfor y = v0yt – (1 / 2)gt2(These equations describe the xand y positions of a projectile that starts at the origin.) Youshould obtain an equation of the form y = ax + bx2 wherea and b are constants
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Max Height = V;*sin's (visine) 28 28 Max Ronge = v? sin 20 Vi Height 8 = 9. 81 m/sz Range 1.) A very small cannon (very small, meaning we can disregard its size) has a muzzle velocity (velocity of the projectile as it leaves the barrel) of 30 m/s and is at an angle of 35° from the ground. a.) What is the Max Height? b.) What is the Range? Somple . Diagrom Vi = inital velocty cmuzzle velocity) 0 = ongle from the grond Vi = 30 m/e Height e= 350 Ronge a) Monx Height : (30) (ain 5) Max Height = V;? sin?e 28 15- 09 m R(9.81) b.) Ronge : v.* sin 20 (30)2 sin (70*) Ronge = 86.21M * Incicode your final answers. In this case, I used a box