derive th solutions for maximum height ,time of descent of a body
Answers
Answer:
(i) Let H be the maximum height reached by the projectile in time t
1
For vertical motion,
The initial velocity = usinθ
The final velocity =0
Acceleration =−g
∴using,v
2
=u
2
+2as
0=u
2
sin
2
θ−2gH
2gH=u
2
sin
2
θ
H=
2g
u
2
sin
2
θ
(ii) Let t, be the time taken by the projectile to reach the maximum height H.
For vertical motion,
initial velocity =usinθ
Final velocity at the maximum height =0
Acceleration a=−g
Using the equation v=u+at
1
0=usinθ−gt
1
gt
1
=usinθ
t
1
=
g
usinθ
Let t
2
be the time of descent.
But t
1
=t
2
i.e. time of ascent= time of descent.
∴ Time of flight T=t
1
+t
2
=2t
1
∴T=
g
2usinθ
(iii) Let R be the range of the projectile in a time T. This is covered by the projectile with a constant velocity ucosθ.
Range=horizontal component of
velocity × Time of flight
i.e, R=ucosθ.T
R=ucosθ.
g
2usinθ
R=
g
u
2
sin2θ
∵2sinθ⋅cosθ=sin2θ
Answer:
for height
The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 θ i 2 g . The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin .
for time
The time of flight is the time taken by a body to remain in air and is given by the sum of the time of ascent (t1) and the time of (t2).
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