Question

.A cannonball is launched upward with a velocity of 73.5 m/s at an angle of 20 degrees above the

horizontal.

(a) How long is the cannonball in the air?

(b) How far away does it land?

() How high does it travel? Cut your time in half!

Answer 1

time of flight=( 2U sinx ) ÷ g

Step-by-step explanation:

a)

u=73.5m/s , x= 20° , g =10m/s^2 then t= {2×73.5 × sin 20°} ÷ 10 = 134.2 ÷ 10 = 13.42 sec b) range is the distance, range= (u^2 sin 2 x ) ÷g = ({73.5 }^2 × sin 2 × 20 )÷ 10 =4025.3÷10 = 402.53meters. I couldn't finish the question so sorry