The range of a Cannon ball fired horizontally from a laboratory table equals to 8/ 3 times the height of the table . what is the direction of the velocity of the projectile as it strikes the ground?

Answer 1

Girma

Answer 2

please work this question ?

Answer 3

range = 8/3 max-height
range = (U^2sin2theta)/g
max height = (u^2sin^2theta)2g
so, (U^2sin2theta)/g = 8/3 (u^2sin^2theta)2g
things that cancel out cancel out like u^2 , g
and left with
sin2theta = 8/3 sin^2thea
and sin2theta = 2sin theta x cos theta from trig properties
and sin^2 theta = sin theta x sin theta
so , 2sin theta x cos theta = 8/3 sin theta x sin theta
things that cancel out again one of the sin theta, multiply both sides by 1/2
cos theta =8/6 sin theta
6/8 = sin theta/cos theta, which is tan theta
tan theta = 3/4
to find the angel just plug the inverse of tan theta in ur calculator it will turn out to be
37 degrees. hope that helps

Answer 4

It's 37degree

Answer 5

We know that this equation is horizontal
So initial vertical velovity is equal to zero and tan theta is equal to v in y direction over v in x direction so to find v in x direction
X=v in x times the time
X=8/3h given
So, v in x=x/t
And we don't have time so we need to find it usin another equation in x or y time is the same so,
Y=ut+1/2at^2. But u is zero so
Y=1/2at^2
t=√2h/g
So when we plug it in to the equation
Vx=x/√2h/g=8/3h×√2h/g
So, we need to find the value of v in y direction so,
V=u+gt, but u is zero, V=gt
V=g×√2h/g
Ao now we have both values so,
Tan theta=8/3h×√2h/g the whole over
g×√2h/ when we divide those values we get tan theta=6h/8h, h cancel out. So the answer is tan invers(3/4) you can check it on your mobile 36.6... Which is rounded to be 37°