Question

pls do 11 ques and surely i will make it brainliest

Answer 1

150 the answer of the question

Answer 2

vcos(60°)=80
v×½=80
v=160m/s
so the horizontal component of the initial velocity will be=vcos(30°)=160×√3/2=80√3m/s
now for the time taken by the ball to reach its maximum height=T/2=t'(say)
so,
u=80m/s(which is vertical component of vel.)
v=0m/s
a=-10m/s²
therefore t'=(0-80)/(-10)=>t'=8sec
so T/4=8/2=4seconds
it's vertical velocity after 4 seconds will be v'(say)
v'=u+at=>v'=80+(-10)4
v'=40m/s
but the horizontal component of the velocity will not change in its whole flight
and will be =80√3m/s
so for the magnitude of the velocity
=√[(40)²+(80√3)²]
=40[√{1+(2√3)²}]
=40[√13]=40×3.6=144m/s which is nearly equal to 150m/s