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3 years ago

How to I isolate t? (time value)

Physics

1 answer:

Vlad1618 [11]3 years ago

4 0

Heya!!!

Answer to your question:

All you need to do is to compare both the equations...
On comparing, you'll get
d=6m
v=6.5m/s
a=-9.81m/s^2
we know,

6= 6.5t -0.5 (9.81)t^2
6=6.5t-4.905 t^2
4.905 t^2-6.5t +6=0
now you can use quadratic eq. to solve this.

Hope it helps ^_^

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A football punter wants to kick the ball so that it is in the air for 4.2 s and lands 55 m from where it was kicked. Assume that

earnstyle [38]

Answer:

$57.24^{\circ}$

24.21 m/s

Explanation:

x = Displacement in x direction = 55 m

y = Displacement in x direction = 0

$y_0$ = Height of the ball when the ball is kicked = 1 m

t = Time taken = 4.2 s

$a_y=g$ = Acceleration due to gravity = $-9.81\ \text{m/s}^2$

u = Initial velocity of ball

Displacement in x direction is given by

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow 55=4.2u_x+0\\\Rightarrow u_x=\dfrac{55}{4.2}\\\Rightarrow u\cos\theta=13.1\ \text{m/s}\\\Rightarrow u=\dfrac{13.1}{\cos\theta}$

Displacement in y direction is given by

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 0=1+4.2u\sin\theta+\dfrac{1}{2}\times (-9.81)\times 4.2^2\\\Rightarrow 0=4.2u\sin\theta-85.5242\\\Rightarrow u\sin\theta=20.36\ \text{m/s}\\\Rightarrow u=\dfrac{20.36}{\sin\theta}$

$\dfrac{13.1}{\cos\theta}=\dfrac{20.36}{\sin\theta}\\\Rightarrow \dfrac{20.36}{13.1}=\dfrac{\sin\theta}{\cos\theta}\\\Rightarrow \theta=\tan^{-1}\dfrac{20.36}{13.1}\\\Rightarrow \theta=57.24^{\circ}$

The ball should be kicked at an angle of $57.24^{\circ}$

$u=\dfrac{13.1}{\cos\theta}=\dfrac{13.1}{\cos57.24^{\circ}}\\\Rightarrow u=24.21\ \text{m/s}$

The initial speed of the ball is 24.21 m/s.

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3 years ago

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tatuchka [14]

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Explanation:

a) $v_{x} = 45 cos40$

b) $v_{y}=45sin40$

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$0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s$

d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.

$x=vt\\x=45cos40(2.95) =102m$

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