Hello help rn or else Ima tell every1 if ur unis that u have cooties
So i tried this qn. I was able to pull out the ans by converting to cartesian form buttttt I tried subbing 21s and the vectors didn't equate so idfk bruh. bc the x-component is the same but the y aint
for the golf ball, a(t) = -10j (we will need to make t -> t - 5 at the end, because we have that t = 0 for the golf ball's launch at 5 seconds)
v(t) = -10tj + C
v(0) = Vi = C
v(t) = Vi -10tj
s(t) = Vti - 5t^2j + C
s(0) = 20j = C
s(t) = Vti + (20-5t^2)j is thus the equation of the stone
now, setting t->t-5, we have that s(t) = V(t-5)i + (20-5(t-5)^2)j
equating the positions j component;
20-5(t-5)^2 = 45t-5t^2
20-5(t^2-10t+25) = 45t-5t^2
20+50t-125=45t (crossed out the squared terms)
5t=105
therefore, t= 21 seconds after the golf ball was hit, is when it will collide with the stone
to find the stone's speed at collision, we equate the x components of position at 21 seconds, which gives V(21-5) = 45sqrt(3)(21)
thus, V(21-5) = 945sqrt(3); and at 21 seconds, the velocity of the stone is V(21-5)i -10(21-5)j = 945sqrt(3)i-160j (we had to let t -> t-5 in v(t))
then the magnitude of that is the speed
