Question

A ball is thrown at t = 0 at an angle, ?i = 53O above the horizontal with an initial velocity v0. The horizontal component of the initial velocity is, v0x = 12 m /s(a) Write the vertical component of the initial velocity, v0y, and also write the velocity and acceleration of the ball at the highest point. (b) What are the horizontal and vertical components of velocity at t = 2 s? (g = 10 m/s2 , sin 53° = 0.8, cos 53° = 0.6).

Name: a ball is thrown at t 0 at an angle i 53o above the horizontal with an initial velocity v0 the horizontal component of the initial velocity is v0x 12 m sa write the vertical component of the initial v
Uploaded: 2021-11-26T04:29:21-08:00
Duration: 6 min 23 s
Channel: Ameer Said
Description: a ball is thrown at t 0 at an angle i 53o above the horizontal with an initial velocity v0 the horizontal component of the initial velocity is v0x 12 m sa write the vertical component of the initial v

          A ball is thrown at t = 0 at an angle, ?i = 53O
above the horizontal with an initial 
velocity v0. The horizontal component of the initial velocity is, v0x = 12 m /s(a) Write the vertical 
component of the initial velocity, v0y, and also write the velocity and acceleration of the ball at the 
highest point. (b) What are the horizontal and vertical components of velocity at t = 2 s? (g = 10 
m/s2
, sin 53° = 0.8, cos 53° = 0.6).

Added by Abdullah A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ameer Said

11/26/2021

Step 1

Substituting the values, we get \(v_{0y} = 20 \times \sin(53^\circ) = 16\) m/s. Show more…

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A ball is thrown at t = 0 at an angle, ?i = 53O above the horizontal with an initial velocity v0. The horizontal component of the initial velocity is, v0x = 12 m /s(a) Write the vertical component of the initial velocity, v0y, and also write the velocity and acceleration of the ball at the highest point. (b) What are the horizontal and vertical components of velocity at t = 2 s? (g = 10 m/s2 , sin 53° = 0.8, cos 53° = 0.6).