Question

sunil played a stroke and hits a ball with a velocity of 20 m/s at an angle of elevation 45 degree.kapil standing near the boundary along the line of stroke,50m away from the batsman,startsrunning to catch the ball at the instant the ball is hit.He managed to run at constant acceleration and at the last moment he took the catch 2 m above the level of the stroke point.Find
1.the acceleration with which kapil must have run.
2.velocity of the ball when caught by fielder
3.velocity of fielder while catching the ball.

please use the normal physics motion formula to solve so that i can understand and if u can give other problems related to this math , i will be grateful.thank you.

Answer 1

Answer on Question #48422-Physics-Mechanics-Kinematics-Dynamics

Sunil played a stroke and hits a ball with a velocity of $20\,\mathrm{m/s}$ at an angle of elevation 45 degree. Kapil standing near the boundary along the line of stroke, $50\,\mathrm{m}$ away from the batsman, starts running to catch the ball at the instant the ball is hit. He managed to run at constant acceleration and at the last moment he took the catch $2\,\mathrm{m}$ above the level of the stroke point. Find

1. the acceleration with which kapil must have run.

2. velocity of the ball when caught by fielder

3. velocity of fielder while catching the ball.

Solution

This is a problem about a projectile motion.

The accelerations in the $x$ and $y$ directions can be integrated to solve for the components of velocity at any time $t$, as follows:

$$a_x = 0; \quad a_y = -g.$$

The accelerations in the $x$ and $y$ directions can be integrated to solve for the components of velocity at any time $t$, as follows:

$$v_x = v_0 \cos \theta, \quad v_y = v_0 \sin \theta - g t.$$

The magnitude of the velocity (under the Pythagorean theorem):

$$v = \sqrt{v_x^2 + v_y^2}.$$

At any time $t$, the projectile's horizontal and vertical displacement:

$$x = v_0 t \cos \theta; \quad y = v_0 t \sin \theta - \frac{g t^2}{2}.$$

We need to find the time of flight of the ball.

$$y(t_f) = v_0 t_f \sin \theta - \frac{g t_f^2}{2}$$

or

$$2\,\mathrm{m} = 20\,\frac{\mathrm{m}}{\mathrm{s}} \cdot t_f \sin 45 - \frac{10\,\frac{\mathrm{m}}{\mathrm{s}^2} \cdot t_f^2}{2} \rightarrow 5 t_f^2 - \frac{20}{\sqrt{2}} t_f + 2 = 0.$$ $$D = \left(\frac{20}{\sqrt{2}}\right)^2 - 4 \cdot 5 \cdot 2 = 160.$$ $$t_f = \frac{\frac{20}{\sqrt{2}} \pm 4\sqrt{10}}{2 \cdot 5} = \sqrt{2} \pm \frac{4}{\sqrt{10}}$$

The time $\sqrt{2} - \frac{4}{\sqrt{10}} \approx 0.14\,\mathrm{s}$ doesn't fulfill the conditions, so

$$t _ {f} = \sqrt {2} + \frac {4}{\sqrt {1 0}} \approx 2. 7 \mathrm {s}.$$

The velocity of the ball when caught by fielder is

$$v _ {y} \left(t _ {f}\right) = 2 0 \frac {m}{s} \cdot \sin 4 5 - 1 0 \frac {m}{s ^ {2}} \cdot 2. 7 \mathrm {s}; v _ {x} = 2 0 \frac {m}{s} \cdot \cos 4 5$$ $$v _ {b} = \sqrt {\left(2 0 \cdot \frac {1}{\sqrt {2}} - 1 0 \cdot 2 . 7\right) ^ {2} + \left(2 0 \cdot \frac {1}{\sqrt {2}}\right) ^ {2}} = 1 9. 1 \frac {m}{s}.$$

The projectile's horizontal and vertical displacement at time $t_f$ :

$$x = 2 0 \cdot \frac {1}{\sqrt {2}} \cdot 2. 7 = 3 8. 2 \mathrm {m}.$$

So, the distance travelled by the catcher:

$$x _ {c} = 5 0 - 3 8. 2 = 1 1. 8 \mathrm {m}.$$

He managed to run at constant acceleration, so

$$x _ {c} = \frac {1}{2} a t _ {f} ^ {2}.$$

The acceleration with which kapil must have run is

$$a = \frac {2 x _ {c}}{t _ {f} ^ {2}} = \frac {2 \cdot 1 1 . 8}{2 . 7 ^ {2}} = 3. 2 \frac {m}{s ^ {2}}.$$

Velocity of fielder while catching the ball is

$$v _ {c} = a t _ {f} = 3. 2 \cdot 2. 7 = 8. 6 \frac {m}{s}.$$

Answer: 1. 3. 2 $\frac{m}{s^2}$; 2. 19. 1 $\frac{m}{s}$; 3. 8. 6 $\frac{m}{s}$.

http://www.AssignmentExpert.com/