February 1, 2008
PHY2053 Discussion
Quiz 3 (Chapter 3.4-3.5 & 4.1-4.4)
Name:
UFID:
*1. (5 pts) A stone is thrown horizontally from the edge of a cliff with a speed of 18.0 m/s.
The cliff is 50.0 m above a flat, horizontal beach. Find the horizontal range of the stone.
We first calculate the time when the stone hits the surface of the beach:
Δy = -(1/2)gt² ⇒ -50 = -0.5×9.8×t² ⇒t = √(50/4.9) = 3.19 s
The horizontal displacement at this time is given by
Δx = 18×3.19 = 57.4 m
***2. (5 pts) In the standard game of darts, a dart is thrown from a point 2.37 m away
form the dartboard and the bullseye is 1.73 m high. Now, a dart is thrown horizontally
toward the dartboard, from an initial height of 1.73 m. It hits the board 0.3 m below the
bullseye. Assuming darts are thrown at the same initial speed, at what angle does a
dart have to be thrown so that it hits the bullseye?
Using the result of the first dart, we express the initial velocity v in terms of the
horizontal distance L = 2.37 m and y displacement h = 0.3 m:
L = vt ⇒ t = L/v
Δy = -(1/2)gt² ⇒ -h = -(1/2)g(L/v)² ⇒ v² = gL²/(2h) (v = √(gL²/(2h)) = 9.58 m/s)
Since the horizontal range is given by
0 = vcosθt-(1/2)gt² ⇒ t = 2vcosθ/g, Δx = vcosθ(2vcosθ/g) = v²sin2θ/g
The dart have to be thrown so that Δx = L, we get
sin2θ = gL/v² = gL/( gL²/(2h)) = 2h/L = 2×0.3/2.37
Thus
2θ = sinˉ¹(0.253) = 14.66, θ = 14.66/2 = 7.33º
**3. (5 pts) A 200 m wide river flows due east at 3.00 m/s. A boat crosses the river from
the south shore to the north shore by maintaining a velocity of 10.0 m/s at an angle of
15º west of north relative to the water. How far upstream or downstream is the boat
when it reaches the north shore?
The velocity of the boat relative to the water and the velocity of the water relative to the
shore are given by
V = (10cos105º, 10sin105º) = (-2.588 m/s, 9.66 m/s)
V’ = (3 m/s, 0 m/s)
Adding these two, we get the velocity of the boat relative to the shore:
V’’ = (-2.588+3, 9.66+0) = (0.412 m/s, 9.66 m/s)
Thus the boat moves in the direction of
θ= tanˉ¹(9.66/0.41) = 87.56º ⇒ 2.44º east of north
The boat is carried downstream by
Δx = 200tan2.44º = 8.52 m
**4. (5 pts) A 0.50 kg ball is dropped from rest at a height of 40.0 m above the ground.
The ball hits the ground and rebounds, reaching a maximum height of 20.0m. If the
contact between ball and ground lasted 2.0 ms, what average force was exerted on the
ball by the ground?
The velocities of the ball just before and just after it hits the ground are given by
v² = 0²+2(-9.8)(-40) ⇒ v = -28.0 m/s (Take the negative root b/c the ball is falling down.)
0² = v’²+2(-9.8)(20) ⇒ v’ = 19.8 m/s
The acceleration is given by
a = Δv/Δt = (19.8-(-28))/(2×10ˉ³) = 2.39×10 m/s²
Since the gravity is much smaller than the force exerted by the ground, we can neglect
the gravity. Thus
F = ma = 0.5×2.39×10 = 1.20×10 N