Question

ball is thrown upward from a point on the side of a hill which slopes upward uniformly at angle 28°.Intial velocity of the ball is v0= 33 m/s and at an angle 65°(with respect to the horizontal. At what distance up the slope the ball strike and in what time?
please answer it quickly​

Answer 1

Answer:

about 30 to

o 45 seconds .

Answer 2

Answer:

suppose this means that the angle of launch on the side of the hill is 65˚. This must mean relative to horizontal, as 65˚ relative to the hillside would be facing down the hill.

Say that [math]t[/math] is the time into the flight in seconds, [math]x[/math] the horizontal distance from the start point, [math]y[/math] the vertical distance from the start point, and [math]s[/math] the distance up the slope.

Considering horizontal and vertical components of the flight separately we have that:

[math]\qquad x=33\cos(65˚)t[/math]

[math]\qquad y=33\sin(65˚)t-\frac{1}{2}gt^2[/math]

Now at the point the ball hits the slope then:

[math]\qquad y=\tan(28˚)x[/math]

[math]\qquad 33\sin(65˚)t-\frac{1}{2}gt^2=33\tan(28˚)\cos(65˚)t[/math]

We’re not interested in the solution at [math]t=0[/math], so we can cancel [math]t[/math] and then solve for the remaining [math]t[/math]:

t=66g(sin(65˚)−tan(28˚)cos(65˚))

We also have that:

s=xcos(28˚)=33cos(65˚)tcos(28˚)

and inserting the equation for t above gives the answer:

s=2178cos(65˚)gcos(28˚)(sin(65˚)−tan(28˚)cos(65˚))

Putting this into the calculator with g=9.8 m/s 2 gives:

s≈711g≈72.5 metres

As a footnote, a slightly simpler and more elegant approach if you’re up for it is to rotate the axes by 28˚ so that the x-axis is along the slope. Then the launch is at 65–28=37˚ and gravity slants back at 28˚ from vertical. This may sound awkward but has the advantage that s=x and the point we want is where y=0 . The equations of motion are then the nice symmetrical:

x=33cos(37˚)t−12gsin(28˚)t2

y=33sin(37˚)t−12gcos(28˚)t2

Setting y=0 and solving for t≠0 gives:

t=66sin(37˚)gcos(28˚)

and then inserting this into the equation for x with a little re-arranging gives the answer:

s=x=2178sin(37˚)gcos(28˚)(cos(37˚)−tan(28˚)sin(37˚))

which of course evaluates to the same value as before.

Finally notice that by using this expression for t in the original expression for s in terms of t we get a simplified expression for the answer:

s=2178sin(37˚)cos(65˚)gcos2(28˚)

So the general solution for slope a , launch angle b and launch speed v is

s=2v2sin(b−a)cos(b)cos2(a)