Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Here, the horizontal component of initial velocity will be vcosq and vertical component of velocity will be vsinq.
The horizontal distance travelled in time t will be
x = (vcosq)t – ½ gt2
The vertical distance travelled in time t will be
x = (vsinq)t – ½ gt2
so, we have
x= (vcosq)(2vsinq/g) + ½ a(2vsinq/g)2
and for maximum horizontal range, we differentiate (1) w.r.t q
dx/dq =cos2q+a/g sin2q =0
(or)
Tan2q = –g/a = –g/QE/m = –mg/QE
Thus, the maximum distance the object can travel horizontally before striking ground is
– mg / QE
Thanks & Regards,
Vasantha Sivaraj,
askIITians faculty
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !