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qaws [65]
3 years ago
15

During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a vel

ocity of 4.8 m/s, grabs and holds onto him so that they move off together with a velocity of 2.5 m/s. If the mass of the tackler is 100 kg, determine the mass of the receiver in kilograms. Assume momentum is conserved.
Physics
1 answer:
zmey [24]3 years ago
4 0

Answer:

Mass of receiver is 92 kg        

Explanation:

We have given mass of tackler m_1=100kg

Let the mass of receiver is m_2kg

When tackler moving alone velocity is v_i=4.8 m/sec

And when tackler and receiver is together velocity is v_f = 2.5 m/sec

So from conservation of momentum

m_1v_i=(m_1+m_2)v_f

100\times 4.8=(100+m_2)\times 2.5

2.5m_2+250=480

2.5m_2=230

m_2=92kg

So mass of receiver is 92 kg

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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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