physics
posted by anon .
We are standing at a distance d=15 m away from a house. The house wall is h=6 m high and the roof has an inclination angle β=30 ∘. We throw a stone with initial speed v0=20 m/s at an angle α= 35 ∘. The gravitational acceleration is g=10 m/s2. (See figure)
(a) At what horizontal distance from the house wall is the stone going to hit the roof  s in the figure? (in meters)
(b) What time does it take the stone to reach the roof? (in seconds)

The time to reach the highest point is
tₒ= 2vₒ•sinα/2g =2•20•sin35/2•10 =1.15 s
The time for covering the distance to the wall ‚d’ is
t₁=d/vₒ•cosα =15/20•cos35= 0.92 s
At horizontal distance d from the initial point the ball is at the height
h₁=vₒ•sinα•t₁ gt₁²/2 =
=20•sin25•0.92 10•0.92²/2 =6.31 m.
The highest position of the ball moving as projectile is
H= vₒ²•sin²α/2g =20²• sin²35/2•10 = 6.38 m.
Therefore, the ball meets the roof at its upward motion, =>
d+s=vₒ•cosα•t …..(1)
h+s•tanβ= vₒ•sinα•t  gt²/2 …..(2)
From (1)
s = vₒ•cosα•t d
Substitute ‘s’ in (2)
h +tanβ(vₒ•cosα•t –d) =
=vₒ•sinα•t  gt²/2,
6+ 0.58(20•0.82•t 15) = 20•0.57•t 5t²,
5t² 1.9t2.7 =0
t=0.95 s.
s= vₒ•cosα•t –d=
=20•0.82•0.95 – 15=
=15.58 – 15 =0.58 m