Question

An object is launched at a velocity of 20m/s in a direction making an angle of 25 degree upward with the horizontal. Find maximum height, time of flight and horizontal range.​

Answer 1

Answer:

• maximum height = 3.64 m
• time of flight = 1.72 sec
• horizontal range = 31.26 m

Explanation:

Given :

An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.

To find :

• the maximum height
• time of flight
• horizontal range

Solution :

here,

u = 20 m/s

θ = 25°

g = 9.8 m/s²

Maximum height :

Maximum height is given by,

  $\boxed{\bf H=\dfrac{u^2sin^2\theta}{2g}}$

Substitute the values,

  $\sf H=\dfrac{20^2 \times sin^2 25^{\circ}}{2(9.8)} \\\\\ \sf H=\dfrac{400 \times 0.178084}{19.6} \\\\ \sf H \simeq 3.64 m$

Time of flight :

Time of flight is given by,

  $\boxed{\bf T=\dfrac{2usin\theta}{g}}$

Substitute the values,

$\sf T=\dfrac{2 \times 20 \times sin25^{\circ}}{9.8} \\\\ \sf T=\dfrac{40 \times 0.422}{9.8} \\\\ \sf T= 1.72 \: sec$

Horizontal Range :

Horizontal Range is given by,

 $\boxed{\bf R=\dfrac{u^2sin2\theta}{g}}$

Substitute the values,

$\sf R=\dfrac{20^2 \times sin2 (25^{\circ})}{9.8} \\\\\ \sf R=\dfrac{400 \times sin 50^{\circ}}{9.8} \\\\ \sf R =\dfrac{400 \times 0.766}{9.8} \\\\ \sf R=31.26 \: m$

Answer 2

Answer:

Given : An object is launched at a velocity of 36m/s in a direction making an angle of 29⁰ upward with the horizontal

To Find :

a) What is the maximum height reached by the object?

b) What is the total flight time (between launch and touching the ground) of the object?

c) What is the horizontal range

c) What is the magnitude of the velocity of the object just before it hits the ground?​

Solution:

Horizontal Velocity =  20cos25°  = 18.13 m/s

Vertical Velocity =      20Sin25°    = 8.45 m/s

Horizontal Velocity  remains constant

Vertical velocity becomes 0 at max height

V = u + at

a = - g  = -9.8 m/s²

=> 0  =8.45 - 9.8t

=> t =0.86 sec

Total Time of flight = 2 * 0.86  = 1.72 sec

V² - U² = 2as

Max height  =  (v² - u²)/2a  =  -(8.45)²/(2*(-9.8))

= 3.64 m

Horizontal range = 1.72 x  18.13   = 31.18 m

magnitude of the velocity of the object just before it hits the ground

= 20 m/s     ( based on conservation of energy )

Initial KE + initial PE  =  Final KE + Final PE

Initial and final PE = 0  

Initial KE =  Final KE

Hence Magnitude of velocity will be same

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Explanation: