A ball is thrown with an initial ve locity of 70 km per second, at an a ngle  of 35° with the horizontal. Find th e vertical and horizontal component s  of the velocity.                ​

i. The horizontal component of the ball = 15.8

ii. The vertical component of the ball = 11.2

Step-by-step explanation:

A thrown ball is an example of a projectile, whose motion can be explained with respect to both horizontal and vertical components.

Horizontal component of the ball = U Cos θ

Vertical component of the ball = U Sin θ

where U is the initial velocity of the ball, and θ is the angle of projection.

Thus for the given question,

U = 70 km/hr (19.44 m/s)

θ = 35°

So that;

i. The horizontal component of the ball = 19.44 Cos 35°

                                               = 19.44 x 0.8192

                                               = 15.785

The horizontal component of the ball = 15.8.

ii. The vertical component of the ball = 19.44 Sin 35°

                                                = 19.44 x 0.5736

                                                = 11.151

The vertical component of the ball = 11.2

found it xd it ok yt

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?