Thank you for registering.

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

Please check your email for login details.
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

A particle is released from a height H= 400m. Due to the wind the particle gathers the horizontal velocity component v x = (sqrt5)y , where y is the vertical displacement of the particle from the point of release, then find: (a) The horizontal drift of the particle when it strikes the ground. (b) the speed with which the particle strikes the ground.

A particle is released from a height H= 400m. Due to the wind the particle gathers the horizontal velocity component vx = (sqrt5)y, where y is the vertical displacement of the particle from the point of release, then find:
(a) The horizontal drift of the particle when it strikes the ground.
(b) the speed with which the particle strikes the ground.

Grade:11

1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
6 years ago
Dear student,
a) The horizontal velocity when it strikes the ground would be equal tov_{x}=\sqrt{5}y=\sqrt{5}\times 400=400\sqrt{5}
b) The vertical velocity would be given by:
v_{y}^{2}=2gH=2\times 10\times 400=8000 \Rightarrow v_{y}=\sqrt{8000}=40\sqrt{5}
Hence, the net velocity would be given by:
v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{\left ( 400\sqrt{5} \right )^{2}+\left ( 40\sqrt{5} \right )^{2}}=\sqrt{800000+8000}=\sqrt{808000}
Regards
Sumit

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free