To find the initial velocity of the ball, the velocity vector must be broken down into Vx and Vy1. In this case, Vx = V cos8 and Vyl the time of flight would be equal to 4/VCos8 in the horizontal direction. The time interval = VSin8. Given that the maximum distance is 4m, would be substituted Vylt + 1/2 at - given 10m, and a = -9.8m/s. After solving for V, the ball initially travels with a velocity of 8.71 m/s @8°. The horizontal velocity is 8.63 m/s [8.71cos8] and since the into the formula dinitial = height dinitial = Om, height range is 4m, the time of flight is 0.463 seconds. The maximum height of the launch can be calculated given that vy2 = 0 m/s, vyl = 8.71sin8 = 1.21 m/s, dy = 0.491 m and a = -9.8 m/s- using the formula hmax = ((vy2? - vy1)/2a) + dy. After calculations, the maximum height of the projectile is 0.566m. Finally, the impact velocity can be calculated by first finding the vy2 using vy2 = Vvyl? + 2ad - given that vyl =- 8.71sin8 =-121m/s, a =- 9.8 m/s and d=- 0.491m. After calculations, vy2 = 3.33 m/s. To find the impact velocity, the horizontal and vertical components of velocity must be added [Vimpact = Vvx? +vy2] to give Vimpact = 9.25 m/s @ 21 ah. Refer to Figure 4 for a diagram of the projectile motion.
To find the initial velocity of the ball, the velocity vector must be broken down into Vx and Vy1. In this case, Vx = V cos8 and Vyl the time of flight would be equal to 4/VCos8 in the horizontal direction. The time interval = VSin8. Given that the maximum distance is 4m, would be substituted Vylt + 1/2 at - given 10m, and a = -9.8m/s. After solving for V, the ball initially travels with a velocity of 8.71 m/s @8°. The horizontal velocity is 8.63 m/s [8.71cos8] and since the into the formula dinitial = height dinitial = Om, height range is 4m, the time of flight is 0.463 seconds. The maximum height of the launch can be calculated given that vy2 = 0 m/s, vyl = 8.71sin8 = 1.21 m/s, dy = 0.491 m and a = -9.8 m/s- using the formula hmax = ((vy2? - vy1)/2a) + dy. After calculations, the maximum height of the projectile is 0.566m. Finally, the impact velocity can be calculated by first finding the vy2 using vy2 = Vvyl? + 2ad - given that vyl =- 8.71sin8 =-121m/s, a =- 9.8 m/s and d=- 0.491m. After calculations, vy2 = 3.33 m/s. To find the impact velocity, the horizontal and vertical components of velocity must be added [Vimpact = Vvx? +vy2] to give Vimpact = 9.25 m/s @ 21 ah. Refer to Figure 4 for a diagram of the projectile motion.
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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How do he do the v impact section part at the bottom? How did he get 9.25 m/s @21 ah
![To find the initial velocity of the ball, the velocity vector must be broken down into Vx and
Vy1. In this case, Vx = V cos8 and Vyl
the time of flight would be equal to 4/VCos8 in the horizontal direction. The time interval
= VSin8. Given that the maximum distance is 4m,
would
be
substituted
height + Vylt + 1/2 at - given
10m, and a = -9.8m/s. After solving for V, the ball initially travels
with a velocity of 8.71 m/s @ 8°. The horizontal velocity is 8.63 m/s [8.71cos8] and since the
into
the
formula
dinitial =
dinitial = Om, height
range is 4m, the time of flight is 0.463 seconds. The maximum height of the launch can be
calculated
given
that vy2 = 0 m/s, vyl
= 8.71sin8 = 1.21 m/s, dy = 0.491 m and
((vy2? - vy1)/2a) + dy. After calculations, the
maximum height of the projectile is 0.566m. Finally, the impact velocity can be calculated
a = -9.8 m/s- using the formula hmax =
by first finding the vy2 using vy2 = Vvyl? + 2ad - given that vyl =- 8.71sin8 =-121m/s,
a =- 9.8 m/s and d=- 0.491m. After calculations, vy2 = 3.33 m/s. To find the impact velocity,
the horizontal and vertical components of velocity must be added [Vimpact = Vvx +vy2 ]
to give Vimpact = 9.25 m/s @ 21° ah. Refer to Figure 4 for a diagram of the projectile motion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0af84c60-49b3-4cee-a0a6-234d88838ea7%2Fa02ce268-2794-4eef-8bc3-ac67ddd00999%2Fbnarzxs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To find the initial velocity of the ball, the velocity vector must be broken down into Vx and
Vy1. In this case, Vx = V cos8 and Vyl
the time of flight would be equal to 4/VCos8 in the horizontal direction. The time interval
= VSin8. Given that the maximum distance is 4m,
would
be
substituted
height + Vylt + 1/2 at - given
10m, and a = -9.8m/s. After solving for V, the ball initially travels
with a velocity of 8.71 m/s @ 8°. The horizontal velocity is 8.63 m/s [8.71cos8] and since the
into
the
formula
dinitial =
dinitial = Om, height
range is 4m, the time of flight is 0.463 seconds. The maximum height of the launch can be
calculated
given
that vy2 = 0 m/s, vyl
= 8.71sin8 = 1.21 m/s, dy = 0.491 m and
((vy2? - vy1)/2a) + dy. After calculations, the
maximum height of the projectile is 0.566m. Finally, the impact velocity can be calculated
a = -9.8 m/s- using the formula hmax =
by first finding the vy2 using vy2 = Vvyl? + 2ad - given that vyl =- 8.71sin8 =-121m/s,
a =- 9.8 m/s and d=- 0.491m. After calculations, vy2 = 3.33 m/s. To find the impact velocity,
the horizontal and vertical components of velocity must be added [Vimpact = Vvx +vy2 ]
to give Vimpact = 9.25 m/s @ 21° ah. Refer to Figure 4 for a diagram of the projectile motion.
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