Posted by anon on .
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, =>
h+s•tanβ= vₒ•sinα•t - gt²/2 …..(2)
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 -1.9t-2.7 =0
s= vₒ•cosα•t –d=
=20•0.82•0.95 – 15=
=15.58 – 15 =0.58 m