Posted by **anon** on Sunday, October 6, 2013 at 9:41am.

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)

- physics -
**KUNOI**, Monday, October 7, 2013 at 6:28pm
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 -1.9t-2.7 =0

t=0.95 s.

s= vₒ•cosα•t –d=

=20•0.82•0.95 – 15=

=15.58 – 15 =0.58 m

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