Answer: 2.04 s
Explanation:
Let the initial velocity be v, Angle of projectile be
Then the horizontal component = v cos θ = 16 m/s
Vertical component of velocity = v sin θ = 20 m/s
Time taken to reach the highest point is half the time taken for total flight.
Time for total flight,


Thus, the football takes 2.04 s to rise to the highest point of its trajectory.
<span>A parent develops a set of rules collaboratively with her child.
Dr. Benjamin Spock believed that children should be treated as individuals.</span>
Answer:
(a) 1.85 m/s
(b) 4.1 m/s
Explanation:
Data
- initial bullet velocity, Vbi = 837 m/s
- wooden block mass, Mw = 820 g
- initial wooden block velocity, Vwi = 0 m/s
- final bullet velocity, Vbf = 467 m/s
(a) From the conservation of momentum:
Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf
Mb*(Vbi - Vbf)/Mw = Vwf
4.1*(837 - 467)/820 = Vwf
Vwf = 1.85 m/s
(b) The speed of the center of mass speed is calculated as follows:
V = Mb/(Mb + Mw) * Vbi
V = 4.1/(4.1 + 820) * 837
V = 4.1 m/s
Answer:
The correct answer would be C, certain display rules are more appropriate than others.
Explanation:
I hope this helps you:)
Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s
,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ +
t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis