An oblique projectile is launched from a point O with an initial velocity u that makes an angle θ with the horizontal. It is observed that the projectile attains a maximum height H at the point $$R2$$,H, where R is the maximum distance from O at which the projectile strikes the ground. Now, if the velocity of the projectile is ν→$$=(20i^+10j^$$ $$)$$ $$ms-1$$ when it is at a height of $$15$$ m above the ground, then based on this information and taking g $$=$$ $$10$$ $$ms-2$$,answer the following questions.The launch speed in $$ms-1$$ isThe value of θ (in$$ $$radian) isThe coordinate, where the maximum height is attained is

Question

An oblique projectile is launched from a point O with an initial velocity u that makes an angle θ with the horizontal. It is observed that the projectile attains a maximum height H at the point $$R2$$,H, where R is the maximum distance from O at which the projectile strikes the ground. Now, if the velocity of the projectile is ν→$$=(20i^+10j^$$ $$)$$ $$ms-1$$ when it is at a height of $$15$$ m above the ground, then based on this  information and taking g $$=$$ $$10$$ $$ms-2$$,answer the following questions.The launch speed in $$ms-1$$ isThe value of θ (in$$ $$radian) isThe coordinate, where the maximum height is attained is

Infinity Learn · Accepted Answer

since v⇀$$=20i^+10j^$$ ⇒$$vx=ux=20ms-1since$$,$$vy=10$$ $$ms-1and$$ $$vy2-uy2=2ayy$$⇒(10)2-uy2=2(-10)(15)⇒$$uy2=400$$⇒  $$uy=20ms-1So$$,$$u=ux2+uy2=202ms-1$$  since $$tan$$ θ$$=uyux=2020=1$$⇒θ$$=450=$$π$$4radianH=uy22g=40020=20$$ $$mR=u2sin(2$$θ$$)g=2022sin(90)10$$⇒$$R=80$$ mNow, the coordinate where the maximum height is attained is $$R2$$,H.so we get (40$$,$$20) m is the answer

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