Answer: the player pushing down on the floor
Explanation:
Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s
The last choice. Two arrows and the arrow up is shorter than the arrow down. Since the guy is falling and he’s opened his chute, he’s slowing down but he’s still falling meaning the force of gravity is stronger than the air resistance.
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