The maximum height attained by a projectile is increased by 10%. Keeping the angle of projection constant,what is percentage increase in time of flight? Ans: 5% HOW?????
Answers
Answer:
5% Approxx
4.88 % ( to be More Precise)
Explanation:
The maximum height attained by a projectile is increased by 10%. Keeping the angle of projection constant,what is percentage increase in time of flight
Let say Velocity = V
Angle of Projection = α
Vertical Velocity = Vsinα
Time to reach max height = Vsinα/g
Time of Flight = 2Vsinα/g
Max height H = V²/2g => V = √(2gH)
=> Time of Flight = 2√(2gH) sinα/g
=> Time of Flight ∝ √H
Now H is increased by 10 %
=> Time of Flight increased by = (1/2)10 = 5% ( approx)
For Exact calculation
Time of Flight T = 2√(2gH) sinα/g With Normal Height
when H is increased by 10 % => H = H + (10/100)H = 1.1H
Tn = 2√(2g*1.1H) sinα/g = √1.1 T
(Tn - T)/T * 100
= (√1.1 T - T )/T * 100
= (√1.1 - 1)*100
= 4.88 %
Answer:
5%
Explanation:
Let the initial velocity of projection be ‘u’, the initial angle of projection ‘θ’ and time of flight be ‘t’.
Initial max. height is, h=
2g
(u
2
sin
2
θ)
Time of flight,t=
g
2usinθ
⇒t
2
=
g
2
4u
2
sin
2
θ
⇒t
2
=(
g
8
)[
2g
(u
2
sin
2
θ)
]=(
g
8
)h
Therefore,
t
2
T
2
=
h
H
Now, new height, H=h+(
100
10
)h=(
10
11
)h
⇒
h
H
=
10
11
And,
t
2
T
2
=
10
11
⇒
t
T
=1.05
⇒
t
T
–1=0.05
⇒(T−t)t=0.05
⇒(T−t)t×100=5%
Thus, the % increase in time of flight is 5%.