2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal velocity of 7.2 m/s. (a) What is the final vertical velocity of the ball? (b) How long does it take to reach the ground? (c) How far does it move horizontally in this time?
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Answer:
Given: Initial height (h₀) = 9.4 m Initial horizontal velocity (v₀x) = 7.2 m/s Acceleration due to gravity (a) = 9.8 m/s²
(a) Find: Final vertical velocity (vₓ)
Formula: The final vertical velocity can be calculated using the formula:
vₓ = v₀y + at
Where: vₓ is the final vertical velocity, v₀y is the initial vertical velocity (which is zero as the shingle is tossed horizontally), a is the acceleration due to gravity, and t is the time.
Solution: Since the initial vertical velocity is zero (v₀y = 0), the final vertical velocity will be equal to the vertical component of the initial velocity:
vₓ = v₀y + at vₓ = 0 + (9.8)(t) vₓ = 9.8t
(b) Find: Time to reach the ground (t)
To calculate the time it takes for the shingle to reach the ground, we can use the equation for vertical motion:
h = h₀ + v₀yt + (1/2)at²
Since the final height (h) is 0, we can rearrange the equation as follows:
0 = h₀ + (v₀y)t + (1/2)at²
Substituting the values:
0 = 9.4 + 0t - (1/2)(9.8)t²
Simplifying:
4.9t² = 9.4
Dividing both sides by 4.9:
t² = 9.4 / 4.9
Taking the square root of both sides:
t = √(9.4 / 4.9)
(c) Find: Horizontal distance traveled (d)
The horizontal distance traveled can be determined using the formula:
d = v₀xt
Where: d is the horizontal distance, v₀x is the initial horizontal velocity, and t is the time.
Substituting the given values:
d = (7.2 m/s) * (√(9.4 / 4.9))
Final Answer: (a) The final vertical velocity of the shingle is 9.8t m/s. (b) It takes √(9.4 / 4.9) seconds for the shingle to reach the ground. (c) The shingle moves a horizontal distance of (7.2 m/s) * (√(9.4 / 4.9)) meters.