# M2 question - projectilesWatch

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#1
there's this m2 question on projectiles which just has me stumpted at the moment... [M2, review exercise 1, question 9]

9. A golf ball is projected with speed 49ms-1 at an angle of elevation θ from a point A on the first floor of a golf driving range. Point A is at a height of 49/15m above horizontal ground. The ball first strikes the ground at a point Q which is at a horizontal distance of 98m from the point A.

a) Show that

6tan2θ - 30tanθ + 5 = 0

b) Hence find, to the nearest degree, the two possible angles of elevation.

c) Find, to the nearest secon, the smallest possible time of direct flight from A to Q.
0
13 years ago
#2
i. Taking horizonatal motion:

s = vt

98 = t49cosθ

t = 2/cosθ

taking vertical motion, a = -g, s = -49/15, t = 2/cosθ, u = 49sinθ:

s = ut - 0.5at2

-49/15 = 49tsinθ - (g/2)t2

-1 = 15tsinθ - 1.5t2

-1 = 30tanθ - 6sec2θ

0 = 30tanθ - 6 - 6tan2θ +1

6tan2θ -30tanθ + 5 = 0
0
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