Question

A ball is projected from the ground at a speed of 10 ms-1 making an angle of 30 degree with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile. The initial height of the second ball(Take g = 10 m/s2) is (a) 6.25 m (b) 2.5 m (c) 3.75

Answer 1

Answer:

Explanation:

t=usin30/g

t=10×1/2/10=1/2 second

H<max>=u^2sin^2 30/2g

= {10×10×1 ÷4} ÷2 ×10

=5\4

h= 1\2× g× t^2

=1/2×10 ×1÷2

=5/4

H<2>= h+H< max>

= 5/4+5/4

=10/4=2.5

Answer 2

Answer:

The initial height of the second ball is equal to 2.5m.

Therefore, the option (b) is correct.

Explanation:

We have given, the first ball projected with speed, u = 10m/s

The angle of projection of ball with ground, θ = 30°

We know that, the acceleration due to gravity, g = 10ms⁻²

Maximum height of the projectile will be:

$h=\frac{u^2sin^2\theta}{2g}$

$h=\frac{(10)^2*sin^2(30^o)}{(2)*(10)}$

$h=\frac{100*(1/2)^2}{20}$

$h=\frac{25}{20}$

$h=1.25m$

The first ball will take  time to reach the maximum height is :

$t=\frac{usin\theta}{g}$

$t=\frac{10*sin30^o}{10}$

$t=0.5sec$

The second ball is released from the point. It will travel a distance in 0.5 seconds.

$S=ut+\frac{1}{2}gt^2$

$S=0+\frac{1}{2} (10)(0.5)^2$

$S=1.25m$

The initial height of the 2nd ball will be the sum of maximum height of first ball and distance travelled by 2nd ball in 0.5second.

Initial height of the second ball = 1.25 + 1.25 = 2.5m

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