Question

please, solve the following problem​

Answer 1

$\bf\huge\underline{\underline{ Question :- }}$

A batter hits a base ball so that it leaves the bat with an initial speed 37m/s at an angle 53∘. Find the position of the ball and the magnitude and direction of its velocity after 2 seconds. Treat the baseball as a projectile.

$\bf\huge\underline{\underline{ Given :- }}$

• u = 37 m/s

• $\theta$ = 53 °

• t = 2 s

$\bf\huge\underline{\underline{ To \: Find:- }}$

• direction of velocity

• magnitude of velocity

$\bf\huge\underline{\underline{ Answer :- }}$

• u = 37 m/s

• $\theta$ = 53 °

• t = 2 s

Vertical Displacement

• $\boxed{\red{\bf s = u_{y} t - \frac{1}{2} gt² }}$

• $\bf s = uSin(\theta ) t - \frac{1}{2} gt²$

• $\bf s = 37×Sin(53 )×2 - \frac{1}{2} ×10×4$

• $\bf\red{ s = 39.5 m}$

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Horizontal Displacement

• $\boxed{\red{\bf s = u_{x} t - \frac{1}{2} gt² }}$

• $\bf s = uCos(\theta ) t - \frac{1}{2} gt²$

• $\bf s = 37×Cos(53 )×2 - \frac{1}{2} ×10×4$

• $\bf\red{s = 44.52 m}$

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∴ So, After 2 Sec the baseball will be lying 39.5m above its point of projection and 44.52m ahead of its point of projection.

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Now , let the vertical component of $\bf v_{y} be v⁰_{y}$ and horizontal component $\bf v_{x} be v⁰_{x}$

• $v⁰_{y} = v_{y} - gt$

• $\bf \red{\implies} v⁰_{y} = vSin(\theta) - gt$

• $\bf \red{\implies} v⁰_{y} = vSin(53) - 10 × 2$

• $\bf \red{\implies} v⁰_{y} = 9.95 m/s$

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• $\bf v⁰_{x} = v_{x} - gt$

• $\bf \red{\implies} v⁰_{x} = vCos(\theta) - gt$

• $\bf \red{\implies} v⁰_{x} = vCos(53) - 10 × 2$

• $\bf \red{\implies} v⁰_{x} = 22.27m/s$

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Magnitude of velocity after 2 second

• $\bf \red{\implies}$ V = $\bf\sqrt{ ( v⁰_{x} )² + (v⁰_{y})² }$

• $\red{\implies}$ V = $\bf\sqrt{ 22.27² + 9.95² }$

• $\bf \red{\implies}$ V =$\bf\red{ 24. 23 m/s }$

• $\bf \theta = tan^{-1} \bigg(\frac{ v⁰_{y}}{v⁰_{x}}\bigg)$

• $\bf \red{\implies} \green{\theta = 23.2 ° }$

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magnitude of velocity = 24.23 m/s

direction = 23.2°