posted by Anonymous 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= 42 ∘. The gravitational acceleration is g=10 m/s2
(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)
8.01 course ah?
did you manage to solve it
yes! Do you have an idea?
no but how did you get it?
@ PHYSICS I didn't get it!
The time to rich 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
@Elena thank you!I found t but s it's wrong
@Elena it works fine I found my mistake !thank you again
you are truly brilliant.
@ Anonymous what do you have for T and S?