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Leona [35]
3 years ago
7

A 75 kg Spider Man running at 3.0 m/s jumps onto a trash can lid that has a mass of 10kg and that is already moving in the same

direction as Spiderman, at 2.0 m/s. What will be the final velocity of Spiderman and the lid assuming that the lid is on a level slick surface and that there is no friction?
Physics
2 answers:
QveST [7]3 years ago
6 0
In order to find the final velocity of the skier and the trash can lid, we may apply the principle of conservation of momentum, which states that the total momentum of a system remains constant. Mathematically, in this case:
m₁v₁ + m₂v₂ = m₃v₃
Where m₃ and v₃ are the combined mass and velocity.

75*3 + 10*2 = (75 + 10)*v₃
v₃ = 2.88 m/s

The final velocity is 2.88 m/s
nikdorinn [45]3 years ago
4 0

The final velocity is 2.88 m/s

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A baseball is batted from a height of 1.09 m with a speed of
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(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

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