Question

A balll is projected with initial speed of 40 m/s making an angle of 30 degrees with the horizontal. Find

(1) Vertical displacement

(2) Horizontal displacement

(3) velocity and angle it is making with horizontal at t=1 and t=3 sec.

Answer 1

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OTHER ANSWERS



Vignesh Ramakrishnan, Aspiring Physicist

Answered Nov 14, 2016 · Author has 366 answers and 432k answer views

Dividing v into horizontal and vertical components , we write two sets of motion equations at t=1.5seconds, assuming gravitational acceleration=10m/s2

Horizontal

v=30(√3/2)+0*1.5=15√3

Vertical

v=30(1/2)-10*1.5=0

Angle wrt horizontal =arctan(0/15√3)=0

Which means the object is perfectly horizontal

14.2k Views · View Upvoters · Answer requested by Anirudh Acharya

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Shivang Khandelwlal, studied Cricket & Books at The Shishukunj International School

Answered Apr 30, 2017

For this question you need to have a basic knowledge of projectile motion.

The time of flight of a vertical projectile is given by:

T= 2u(sin theta)/ g

So by this we get the time of flight equal to 3 seconds.

And as we know time of ascent = time of descent

So time taken to reach the max height where the angle of projectile becomes 0 with respect to horizontal.

So after 1.5 seconds the angle with the horizontal of projectile is 0 degrees.

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Vikas Sangwan, B.E from Army Institute of Technology (2018)

Answered Nov 14, 2016 · Author has 184 answers and 145k answer views

As there is no gravitational force in horizontal direction so ucos30=30×√3/2=25.7m/s, will remain same.

But as there is gravitational force in vertical direction so vertical component will change

And here initial u=usin30=30×1/2=15m/s

v=u+a×t

v= 15+(-10×1.5)=0

So after 1.5 second of flight object will be parallel to horizontal so angle will be 0° will horizontal.