Pls help in this question
The equation of trajectory of a projectile is given as
2
2
2 x
y x . The maximum height of projectile is
(Symbols have usual meanings and all are in SI unit)
Answers
Answer:
if you are not familiar with math, we have another option too.
you know standard trajectory of projectile motion, y = tanα.x - gx²/2u²cos²α
on comparing with y = 2x - x²/2
we get, tanα = 2 ⇒cosα = 1/√5 and sinα = 2/√5
g/2u²cos²α = 1/2
or, g = u²(1/√5)²
or, 10 × 5 = u² ⇒u² = 50
now, maximum height , H = u²sin²α/2g
= {50 × (2/√5)²}/2(10)
= {50 × 4/5}/20
= 40/20 = 2m
hence, maximum height = 2m
Explanation:
Answer:ANSWER
(x
2
+y
2
)
2
=xy
x
4
+2x
2
y
2
+y
4
−xy=0
Differentiating w.r.t x,
4x
3
+2[x
2
.2y(
dx
dy
)+y
2
(2x)]+4y
3
(
dx
dy
)−[
dx
xdy
+y(1)]=0
(
dx
dy
)[4x
2
y+4y
3
−x]+4x
3
+4xy
2
−y=0
(
dx
dy
)[4x
2
y+4y
3
−x]=y−4x
3
−4xy
2
(
dx
dy
)=
4x
2
y+4y
3
−x
y−4x
3
−4xy
2
ANSWER
(x
2
+y
2
)
2
=xy
x
4
+2x
2
y
2
+y
4
−xy=0
Differentiating w.r.t x,
4x
3
+2[x
2
.2y(
dx
dy
)+y
2
(2x)]+4y
3
(
dx
dy
)−[
dx
xdy
+y(1)]=0
(
dx
dy
)[4x
2
y+4y
3
−x]+4x
3
+4xy
2
−y=0
(
dx
dy
)[4x
2
y+4y
3
−x]=y−4x
3
−4xy
2
(
dx
dy
)=
4x
2
y+4y
3
−x
y−4x
3
−4xy
2
Explanation: