Question

Q.75. A body is thrown with a velocity of 20 m/s at an angle of 45° with the horizontal.
a) What is the time taken by the body to reach the ground?
b) What is the maximum height reached by the body?
c) How far from the starting point does it reach the ground?
Take g = 10 m/s2. (2.82m; 10 m; 40 m)

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Answer 1

Answer:

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Answer 2

Given:

• Initial velocity of body, u = 20 ms⁻¹
• Angle of projection, θ = 45°
• Acceleration due to gravity to be taken, g = 10 ms⁻²

To find:

• Time taken by body to reach the ground, T =?
• Maximum height reached by the body, H =?
• Horizontal distance between the starting point and end point of body, R =?

Formula required:

• Formula for time of flight ( Time taken by projectile to reach the ground )

  $\red{\bigstar}\boxed{\sf{T=\dfrac{2\;u\;sin\theta}{g}}}$

• Formula for Maximum height ( Maximum height reached by the body )

$\red{\bigstar}\boxed{\sf{H=\dfrac{u^2\;sin^2\theta}{2\;g}}}$

• Formula for Horizontal range ( Horizontal distance between starting and end point of projectile )

$\red{\bigstar}\boxed{\sf{R=\dfrac{u^2\;sin\;2 \theta}{g}}}$

[ Where T is time of flight, u is initial velocity of body, θ is angle of projection, g is acceleration due to gravity, H is maximum height and R is Range of projectile ]

Solution :

Calculating Time of flight of body

$\implies\sf{ T = \dfrac{2\;u\;sin\theta}{g}}$

$\implies\sf{ T = \dfrac{2\times 20 \times sin\;45^{\circ}}{10}}$

$\implies\sf{ T = 4\times sin\;45^{\circ}}$

$\implies\sf{ T = 4\times\dfrac{1}{\sqrt2}}$

$\implies\underline{\underline{\large{\red{\sf{ T =2.82\;\;s}}}}}$

Calculating Maximum height of projectile

$\implies\sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g}}$

$\implies\sf{ H = \dfrac{(20)^2\times (sin\;45^{\circ})^2}{2 \times 10}}$

$\implies\sf{ H = \dfrac{400\times (sin\;45^{\circ})^2}{20}}$

$\implies\sf{ H = 20 \times (sin\;45^{\circ})^2}$

$\implies\sf{ H = 20 \times \left(\dfrac{1}{\sqrt2}\right)^2}$

$\implies\underline{\underline{\large{\red{\sf{ H =10 \;\;m}}}}}$

Calculating Horizontal range of projectile

$\implies\sf{ R = \dfrac{u^2\;sin\;2\theta}{g}}$

$\implies\sf{ R = \dfrac{(20)^2\times\;sin\;2(45^{\circ})}{10}}$

$\implies\sf{ R = \dfrac{400\times\;sin\;90^{\circ}}{10}}$

$\implies\sf{ R = 40 \times sin\;90^{\circ}}$

$\implies\sf{ R = 40 \times 1}$

$\implies\underline{\underline{\large{\red{\sf{ R = 40\;\;m}}}}}$

Therefore,

• Time taken by body to reach the ground = 2.82 seconds
• Maximum height reached by the body = 10 metres
• Horizontal distance between starting and end point of body = 40 metres.