×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Find time of flight of projectile thrown horizontally with speed 10 metres per secon  from a long inclined plane whichmakes an angle of  = 45 the time of flight is  : please explain with diagram

2 years ago

Susmita
425 Points

A|'''''''''|B
C """"" D
Draw a straight line connecting A and D.The angle ADC is 45°.
The paricle is thrown from point A in the direction AB(horizontally).So at t=0 or initially the particle has only horizontal component of velocity (say v)which is 10m/s and at t=0 the vertical component of velocity(say u) is zero.
Lets say AC=h and after a time t the particle lands on ground.So t is time of flight.
h=ut+(1/2)gt2
or,h=gt2/2=10t2/2=5t2
Now we have to calculate height h.Lets say CD=x.
So tan45°=h/x
or,1=h/x
Or,h=x
Now x=vt=10t
So using equation h=x,
5t2=10t
or,t=2s
2 years ago
Navya khandelwal
29 Points

Time for flight is 2 u sin  theta/g
S for inclined plane t=2u sin theta/g
T=2*10 sin 45/ g sin 45
T=2sec

2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

Post Question