Respuesta :
Answer:
(a) [tex]t=\frac{H}{v_0}[/tex]
(b) [tex]H=\frac{v_0^2}{g}[/tex]
Explanation:
Let the two balls collide at a height x from the ground. Therefore, ball 2 travels a distance of (H-x) before colliding with ball 1.
Using the following Newton's law of motion,
[tex]S=ut+\frac{1}{2}at^2[/tex]
where,
[tex]S[/tex] = displacement
[tex]u[/tex] = initial velocity
[tex]a[/tex] = acceleration
[tex]t[/tex] = time
we can write the equations of motion of the two balls(ball 1 and ball 2 respectively):
[tex]x=v_0t-\frac{1}{2}gt^2[/tex] ......(1) ([tex]a=-g[/tex], ball is moving against gravity)
[tex]H-x=\frac{1}{2} gt^2[/tex] .......(2) (initial velocity is zero; [tex]a=+g[/tex])
Substituting [tex]x[/tex] from equation (1) in (2),
[tex]H-v_0t+\frac{1}{2}gt^2=\frac{1}{2}gt^2[/tex]
or, [tex]t=\frac{H}{v_0}[/tex] ......(a)
(b) Now, it is said that the collision will occur when ball 1 is at it's highest point. That is, it's final velocity must be zero.
This time we shall have to use another equation of motion given by,
[tex]v^2=u^2+2aS[/tex]
where, [tex]v[/tex] = final velocity
therefore, we get for ball 1,
[tex]0=v_0^2-2gx[/tex] ([tex]u=v_0,v=0,a=-g[/tex])
or, [tex]x=\frac{v_0^2}{2g}[/tex]
Putting the value of [tex]x[/tex] in equation (2) and rearranging, we get,
[tex]\frac{g}{2v_0^2}H^2-H+\frac{v_0^2}{2g}=0[/tex]
which is a quadratic equation, whose solution is given by,
[tex]H=\frac{+1\pm\sqrt{(-1)^2-(4\times\frac{g}{2v_0^2} \times\frac{v_0^2}{2g}) } }{2\times\frac{g}{2v_0^2} }[/tex]
[tex]=\frac{v_0^2}{g}[/tex]
(a) The time at which the balls collide is H/[tex]v_{0}[/tex]
(b) The height H is equal to [tex]\frac{v_{0} ^{2} }{g}[/tex]
Let the balls collide at a height x above the ground.
Then the distance traveled by the ball thrown above is x.
And the distance traveled by the ball dropped from height H is (H-x).
(i) Both the balls will take the same time to travel respective distances in order to collide.
[tex]H-x=\frac{1}{2}gt^{2}[/tex]
[tex]x = v_{0}t - \frac{1}{2}gt^{2}[/tex]
We get:
[tex]x=v_{0}t-(H-x)[/tex]
[tex]t=\frac{H}{v_{0}}[/tex] , is the time after which the balls collide.
(ii) Let the ball thrown up attains its maximum height x at the time of thecollision
[tex]v^{2} = u^{2}-2gx[/tex] here v is the final velocity which is 0 when the ball attains maximum height
[tex]0=v_{0} ^{2}-2gx[/tex]
[tex]x=\frac{v_{0} ^{2} }{2g}[/tex] is the maximum height attained.
Now, the ball thrown downward travels distance (H-x) just before collision:
[tex]H-x=\frac{1}{2}gt^{2}[/tex]
[tex]H-\frac{v_{0} ^{2} }{2g}=\frac{1}{2}g\frac{H^{2} }{v_{0} ^{2} }[/tex]
Solving the quadratic equation we get:
[tex]H=\frac{v_{0} ^{2} }{g}[/tex]
Learn more about equations of motion:
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