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3 years ago

10

A quarterback throws a pass at an angle of 35◦ above the horizontal with an initial speed of 25 m/s. the receiver catches the ba

ll 2.55 s later. determine the distance the ball was thrown.

1 answer:

enyata [817]3 years ago

7 0

As this is the projectile motion means the motion of object under constant acceleration, so we use all eq. of motion under constant acceleration.

Angle above horizontal=θ= 35°
Initial speed= v₁ = 25 m/s
Time of flight= t= 2.55 s
Now,
from eq. of motion
x=x₁ + (v₁*cosθ)(t)+1/2 *a*t²
As, there is no acceleration in horizontal direction so a=0 and also initial displacement x₁=0
x= 25*cos(35)*2.55
x=52.22 m

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3 years ago

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as

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Answer:

a) $v=5.6725\,m.s^{-1}$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^{-2}$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^{-1}$

putting the values in eq. (1)

$v^2=0^2+2\times 3.62 \times 10$

$v=8.5088\,m.s^{-1}$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20\times 8.5088$

$p=170.1764\,kg.m.s^{-1}$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)\times v'=p$

$(20+10)\times v'=170.1764$

$v'=5.6725\,m.s^{-1}$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$\frac{1}{2} (M+m).v'^2=(M+m).g.h$

$\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h$

$h\approx 1.6420\,m$

above the normal hanging position.

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