a shot is fired with a velocity 100m/s in a direction making an angle 60 degree with vertical. calculate its time flight, maximum height reached and its horizontal range

#Horizontal component of 100m/s=100 sin60=50×(root 3)=50×1.732=86.6m/s

#Vertical component of 100m/s=100 cos 60°=100/2=50m/s

(¡) Maximum height it can reach=(v^2+u^2) /2a=0+(50^2) / 2×(-10)=125m

(¡¡) The time takes to reach the maximum height =2S/(u-v)=125×2/(50-0)=5s

(¡¡¡) Horizontal range=ut=86.6×5=433m

Explanation:

# At the highest point (vertical) velocity is zero

#Velocity (horizontal) remains constant

(¡)By v^2=u^2+2as equation

(¡¡)By S=1/2(u+v)t equation

(¡¡¡) S=ut therefore horizontal range=ut

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