A ball is launched at an initial velocity of 40m/s at an angle of 40º above the horizontal...?

Question

A ball is launched at an initial velocity of 40m/s at an angle of 40º above the horizontal...

Find the initial velocity in the x direction.

STEPS PLEASE!!! I NEED TO LEARN HOW TO DO THIS.

Answer 1

This is fairly easily solved, since using projectile motion, the velocity in the x direction is always constant and is given by

v_x = v cos φ, where φ is the launch angle with respect to the horizontal.

v_x = 40 cos 40°
v_x = 30.6 m/s

Answer 2

The ball's velocity of 40m/s is composed of two components: X and Y.
If the launch was at 40° then the X velocity is 40m/s*Cos(40°)=30.64m/s and the Y velocity is 40m/s*Sin(40°)=25.71m/s. The sqrt(Vx^2+Vy^2) should be 40m/s.

Answer 3

when deal in 2 space, there is an x compenent and y component. to find each component we need to use the laws of trip for a right angle and make use of cosine and sine.

cos x = a/ y, where y is hypot and a is adjacent leg,s o a = ycosx.

therefore, the x component of velocity will be 40cos40
which equals approx 30.64.

so the x compnent of velocity is 30.64 m.s

and if you had wanted the y component of velocity, it would have been 40sin40.

Answer 4

The vector velocity (V) is 40m/s. and φ is 40°
The x component is Vcos(φ) = 40cos(40)
The y component is Vsin(φ) = 40sin(40)

HTH

Doug