Question : A ball of mass m is thrown vertically upwards. Another ball of mass 2m is thrown up making an angle θ with the vertical. Both of them stay in air for the same time. Find the ratio of the heights attained by the balls.
Doubt by Jaskirat
Solution :
The time of flight and heights attained by the two balls are independent of their masses because there is no mass (m) in the expression of Time of flight and height attained by them.
Case: 1
The ball of mass m is thrown vertically upward.
Let the ball is projected upward with initial velocity u1.
At the highest point, the final velocity of the body will be zero, i.e. v1=0 m/s.
Now Using
v2-u2=2as
(0)2-u12 = 2(-g)H1
-u12=-2gH1
u12=2gH1
v2-u2=2as
(0)2-u12 = 2(-g)H1
-u12=-2gH1
u12=2gH1
H1=u12/2g — (1)
Also
v=u+at
0=u1+(-g)t
0=u1+(-g)t
-u1=-gt
u1=gt
u1=gt
u1/g=t
t=u1/g
t=u1/g
We know,
Time of ascent = Time of descent
Time of ascent = Time of descent
so
Total time of flight (T1) = 2t
T1=2t
T1=2u1/g — (2)
T1=2t
T1=2u1/g — (2)
Case : 2
When the ball of mass 2m is thrown up making an angle θ with the vertical.
Let u2 be the initial velocity, and 90-θ is the angle the projectile makes horizontally.
Let u2 be the initial velocity, and 90-θ is the angle the projectile makes horizontally.
Time of flight (T2)
T2=2u2sin(90-θ)/g
T2=2u2cosθ/g — (3)
T2=2u2cosθ/g — (3)
ATQ
T1=T2
From eq (2) and eq (3)
From eq (2) and eq (3)
2u1/g=2u2cosθ/g
u1=u2cosθ — (4)
Now Height attained by the projectile
H2=u22sin2(90-θ)/g
H2=u22cos2θ/g
H2=(u2cosθ)2/g
H2=u12/g [∵ Using equation (4)]
H2=u12/g — (5)
Dividing equation (1) by (5)
H2=u22cos2θ/g
H2=(u2cosθ)2/g
H2=u12/g [∵ Using equation (4)]
H2=u12/g — (5)
Dividing equation (1) by (5)
H1/H2 = u1/2g/u1/2g
H1:H2=1:1
H1:H2=1:1