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How would I solve this equation for t?sy=u sin(x)t+1/2 ayt2

Question

How would I solve this equation for t?

sy=u sin(x)t+1/2 ayt2

check_circleAnswer
Step 1

Given:

sy = u sin(x) t + 1/2 ay t2 

Step 2

Showing how to solve the given equation for ‘t’...

s usin(x)t-
...(1)
=
The given equation 1 shows the displacement of a particle at time 't
can be arranged into a quadratic equarion
The given equaion
2s, 2usin (x)ta,t
a2usin (x)t2s, 0 ....(2)
+
Comparing the equation 2 with a standard quadratic equation, at bt+c 0
a a,t, b= 2usin(x) and c = -2s,
-btvb2-4ac
Root of the standard quadratic equarion in 't', t =
2a
Putting values of a,b and c
-2u sin xt(2usinx)*-4xa,tx(-2s,)
2(a,r)
Solution for equation 2, t
-2u sin xt2usin x) + 2a,s,t
2(a,t)
-usin xtusin2 x+2a,s,t
.(3)
t=
By putting the required values, value of 't' can be known from equation 3
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s usin(x)t- ...(1) = The given equation 1 shows the displacement of a particle at time 't can be arranged into a quadratic equarion The given equaion 2s, 2usin (x)ta,t a2usin (x)t2s, 0 ....(2) + Comparing the equation 2 with a standard quadratic equation, at bt+c 0 a a,t, b= 2usin(x) and c = -2s, -btvb2-4ac Root of the standard quadratic equarion in 't', t = 2a Putting values of a,b and c -2u sin xt(2usinx)*-4xa,tx(-2s,) 2(a,r) Solution for equation 2, t -2u sin xt2usin x) + 2a,s,t 2(a,t) -usin xtusin2 x+2a,s,t .(3) t= By putting the required values, value of 't' can be known from equation 3

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Physics

Kinematics

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