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2 years ago

13

A tennis player strikes a tennis ball from underneath with her racket. The ball is sent straight up with an initial velocity of

19 m/s. How high above the striking point will the ball travel?
A. 9 m
B. 9.8 m
C. 18 m
D. 19 m

1 answer:

Stels [109]2 years ago

6 0

So the acceleration of gravity is 9.8 m/s so that’s how quickly it will accelerate downwards. You can use a kinematic equation to determine your answer. We know that initial velocity was 19 m/s, final velocity must be 0 m/s because it’s at the very top, and the acceleration is -9.8 m/s. You can then use this equation:

Vf^2=Vo^2+2ax

Plugging in values:

361=19.6x

X=18 m

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Read 2 more answers

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

2 years ago

If the window is 11 m above the ground, find the time the stone is in flight.

rjkz [21]

d=vi*t+(1/2)gt²

d=11 m
g=9.8 m/s²
vi=0 m/s

11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s

Answer: the time the sone is in flight is 1.5 s

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