[ALI ALZAHRANI]

the army fired artillery ls at the enemy 1200 m from the coast of the city , if the speed of the missile 108.4m/s a) what is the angle that must be launched by the l to reach to the enemy.. b) What is the time required for that ..

من قوانين المقذوفات، نستخدم قانون المدى

R=v0^2 sin2Q/g

ومنها نجد أن

sin2Q= R g/v0^2 =1200*9.8/108.4^2=1

يعني أن

2Q= sin^-1(1) = 90

Q= 45

أما الزمن اللازم فيمكن حسابه من العلاقة

t=2 v0 sinQ/g = 2*108.4 sin45/9.8=15.6 s

A l has been fired at speeds of 15 m/s and with an angle of 45ْ with the horizontal. It will reach a maximum height of ;

اقصى ارتفاع يساوي


H=v0^2 sin^2Q/2g=15^2*(sin45)^2/(2*9.8)=5.74 m

A goalkeeper kicked the ball to another player with angle of 30ْ with the horizontal , if the player 45m away , the speed which the goalkeeper kicked the ball with is :

نستخدم قانون المدى

R=v0^2 sin2Q/g

ومنها نجد أن

v0^2= R g/sin2Q =45*9.8/sin60=509.2

وبحساب الجذر التربيعي نجد ان السرعة

v0=22.6 m/s