A ball A is projected from origin with an initial velocity v 0 = 700 cm / s , in a direction 3 7 ∘ above the horizontal as shown in fig. Another ball B 300 cm from origin on a line 3 7 ∘ above the h

Question

A ball A is projected from origin with an initial velocity $v_{0} = 700 cm / s$ , in a direction $3 7^{\circ}$ above the horizontal as shown in fig. Another ball B 300 cm from origin on a line $3 7^{\circ}$ above the horizontal is released from rest at the instant A starts. then how far will $B$ have fallen when it is hit by $A -$

TestHub · Accepted Answer

To solve this problem, we can analyze the motion of ball A and ball B.

Given:

Initial velocity of ball A, $v_{0} = 700 c m / s$

Projection angle for A, $θ = 3 7^{\circ}$

Distance of B from origin along the $3 7^{\circ}$ line, $d = 300 c m$

Acceleration due to gravity, $g = 9.8 m / s^{2} = 980 c m / s^{2}$

Let's consider a coordinate system rotated such that the x-axis is along the line $3 7^{\circ}$ above the horizontal and the y-axis is perpendicular to it.

In this rotated coordinate system:

The initial velocity of A has only an x-component: $v_{A x} = v_{0} = 700 c m / s$ and $v_{A y} = 0$ .

The acceleration due to gravity $g$ has components:

$g_{x} = - g sin (3 7^{\circ})$

$g_{y} = - g cos (3 7^{\circ})$

Ball B is at rest at $(300, 0)$ in this rotated frame and is released. Its initial velocity is $v_{B x} = 0$ , $v_{B y} = 0$ .

The acceleration of B is due to gravity: $a_{B x} = g_{x}$ and $a_{B y} = g_{y}$ .

For ball A:

$x_{A} (t) = v_{0} t + \frac{1}{2} g_{x} t^{2}$

$y_{A} (t) = \frac{1}{2} g_{y} t^{2}$

For ball B:

$x_{B} (t) = d + \frac{1}{2} g_{x} t^{2}$

$y_{B} (t) = \frac{1}{2} g_{y} t^{2}$

When A hits B, their positions must be the same: $x_{A} (t) = x_{B} (t)$ and $y_{A} (t) = y_{B} (t)$ .

From $x_{A} (t) = x_{B} (t)$ :

$v_{0} t + \frac{1}{2} g_{x} t^{2} = d + \frac{1}{2} g_{x} t^{2}$

$v_{0} t = d$

$t = \frac{d}{v _{0}} = \frac{300 c m}{700 c m / s} = \frac{3}{7} s$

Now we need to find how far B has fallen. This is the vertical displacement of B in the original horizontal-vertical coordinate system.

Let's use the original coordinate system.

The position of B at time $t$ is given by:

$x_{B} (t) = d cos (3 7^{\circ})$ (constant horizontal position relative to origin of B's initial position)

$y_{B} (t) = d sin (3 7^{\circ}) - \frac{1}{2} g t^{2}$ (vertical position of B)

The initial height of B is $h_{0} = d sin (3 7^{\circ})$ .

The height of B at time $t$ is $h_{B} (t) = h_{0} - \frac{1}{2} g t^{2}$ .

The distance B has fallen is $Δ h_{B} = h_{0} - h_{B} (t) = \frac{1}{2} g t^{2}$ .

Substitute the value of $t$ :

$Δ h_{B} = \frac{1}{2} (980 c m / s^{2}) (\frac{3}{7} s)^{2}$

$Δ h_{B} = \frac{1}{2} (980) (\frac{9}{49})$

$Δ h_{B} = 490 \times \frac{9}{49}$

$Δ h_{B} = 10 \times 9$

$Δ h_{B} = 90 c m$

The distance B has fallen when it is hit by A is $90 c m$ .

The final answer is $90 c m$ .

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