Grade 11Mechanics

find time of flight of projectile thrown horizontally with speed 50m/s from a long inclined plane which makes an angle of 45 from horizontal.

Akshat

10 Years agoGrade 11

3 Answers

Paras Verma

10 Years ago

Horizontal distance travelled equals vertical distance travelled

Horizintal motion-

accleration=0

velocity=50m/s(constant)

vetical distance=x(say)

speed=distance/time

time=x/50

t²=x²/2500-------Eqⁿ(1)

Vertical motion-

acclⁿ=g

velocity(initial)=0

s=ut+gt²/2

x=10t²/2

t²=2x/10--------Eqⁿ(2)

from Eqⁿ(1) and Eqⁿ(2)

x²/2500=2x/10

x/2500=1/5

x=2500/5=500

Now t=x/50

t=500/50=10sec

Trisha Barui

8 Years ago

Dear Akshat,

Please find the correct solution.

T= 2u sin theeta/g [formula for time o flight]

so t = 2*50* sin 45/10

so t= 10/ root 2

so t = 5 root 2

please ask if any doubt.

Regards,

Arpit

6 Years ago

C orrect answer is root 2.

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find time of flight of projectile thrown horizontally with speed 50m/s from a long inclined plane which makes an angle of 45 from horizontal.

Akshat

3 Answers

Paras Verma

Trisha Barui

Arpit

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