Question

a projectile thrown at an angle of 45 degree to the horizontal has a range of 19.6 its velocity at the highest point is​

Answer 1

Answer:14m/s

Explanation:use range formula

Answer 2

Answer:10m/s

Explanation:theta = 45degrees , range =19.6m , vx [max] at highest pt. =?

firstly find velocity,

by range formula=

R= u^2* sin2theta / g

19.6 = u^2* sin[90] /10                         [g=10m/s^2]

u^2 = 19.6*10         [ sin 90 =1]

u^2 =196    

u=14m/s

then maximum velocity at x co - ordinate is v=ucostheta

hence , 14=  u*cos 45

u =  14*1/root2 = 10m/s

ANS- vel. at highest pt. is 10m/s