Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Range of projectile motion =500mtime=10secondfind maximum height?
500 = u^2*sin(2theta)/g................(1)and10 = u^2*(sin(theta))^2/2g..................(2)solving both eqns. we get,ucos(theta) = 50 andusin(theta) = 50we get u = root(5000)Maximum height = u^2/2g = > 5000/20 = > 250 meter.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -