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 -1.9t-2.7 =0
t=0.95 s.
s= vₒ•cosα•t –d=
=20•0.82•0.95 – 15=
=15.58 – 15 =0.58 m
To find the horizontal distance from the house wall where the stone hits the roof and the time it takes for the stone to reach the roof, we can use the equations of projectile motion.
(a) To find the horizontal distance, we need to analyze the horizontal and vertical components of the stone's motion separately.
The horizontal component of the stone's velocity remains constant throughout its motion. We can find the horizontal distance using the equation:
d = v₀ * cos(α) * t
Given:
v₀ = 20 m/s (initial speed)
α = 35° (launch angle)
d = 15 m (distance from the house wall)
We want to find t, the time it takes for the stone to reach the roof. Rearranging the equation, we get:
t = d / (v₀ * cos(α))
Substituting the given values:
t = 15 m / (20 m/s * cos(35°))
Calculating the value of t, we get:
t ≈ 1.375 s
(b) To find the time it takes for the stone to reach the roof, we can use the vertical motion equation for projectile motion:
h = v₀ * sin(α) * t - (1/2) * g * t²
In this case, h is the height of the roof:
h = 6 m
We want to find t. Rearranging the equation and substituting the given values, we get a quadratic equation in t:
(1/2) * g * t² - v₀ * sin(α) * t + h = 0
Substituting the given values:
(1/2) * 10 m/s² * t² - 20 m/s * sin(35°) * t + 6 m = 0
To solve this quadratic equation, we can either factorize it or use the quadratic formula. Solving the equation using the quadratic formula, we get two values of t:
t ≈ 1.908 s and t ≈ 0.094 s
Since we are interested in the time it takes for the stone to reach the roof, we can discard the value of t ≈ 0.094 s.
Therefore, the stone takes approximately 1.908 seconds to reach the roof.