Physics, asked by braing33, 2 months ago

A projectile is fired with initial velocity 95 m/s at an angle of 50 degrees from the horizontal. After 5 s, it strikes the top of a hill. What is the elevation of the hill above the point of firing?

Answers

Answered by shalinithore100
3

Explanation:

Calculating the time taken by the projectile to reach the ground:

s=ut+

2

1

gt

2

500m=(100m/s)t+

2

1

(9.8m/s

2

)t

2

500=100t+4.9t

2

t=4.15s

Calculating the distance of the target from the hil:

Range = ut=(100m/s)(4.15s)=415m

Calculating the velocity with which the projectile hits the ground:

v=100m/s+(9.8m/s

2

)(4.15s)

2

=268.78m/s

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Answered by KajalBarad
0

21.27 degrees

Given:

A projectile is fired with initial velocity 95 m/s at an angle of 50 degrees from the horizontal

To find:

elevation of the hill above the point of firing

Solution:

Vertical acceleration a = -9.8 m/s^2.

Launch speed S = 95 m/s at 50° elevation.

As function of time, vertical velocity v = u + a*t

where, initial value u = Ssin50 = 72.77 m/s.

To reach apex where v = 0,

elapsed time, t = -u/a = -72.77/-9.8 = 7.43 s.

But for 5 s flight, projectile never reaches apex, so impact time T = 5 s.

At impact, height = uT+ 1/2aT^2

= 72.77*5 -9.8/2*25 = 363.9 -122.5

= 241.4 m.

Horizontal range = S*cos50*T = 305.3 m.

Line of sight displacement

= \sqrt{X^2+Y^2} = 389.2 m.

At impact, vertical velocity v = u + a*T

= 72.77 -9.8*5 = 23.77 m/s

Horizontal velocity w = S*cos50 = 61.06 m/s

Impact speed = \sqrt{w^2 + v^2}

= \sqrt{61.06^2 + 23.77^2}

= 65.53 m/s.

Impact elevation angle = arctan(v/w)

= arctan(23.77/61.06)

= 21.27°

At impact after 5 seconds:

height 241.4m, range 305.3m, displacement 389.2m, velocity 65.53 m/s at 21.27 degrees, still rising.

#SPJ2

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