Question

a projectile is thrown at an angle of 30° with a velocity of 10 m/s the change in velocity during the time interval in which it reaches the highest point is

Answer 1

Tan 30= x/980

x= 565.8 m/s

v= u + at

0= 980 + -9.81t

9.81t= 980

t= 100 s

Answer 2

ta=v-u

And t=usintheta/g=0.5

So v-u=0.5×10=5m/s