A ball was thrown outside the window at an angle of 20° below the horizontal with an initial velocity of 8 m/s. If it lands 3 sec later, how far from the base of the building did it land?
Let ball is thrown with initial velocity 'V'=8 m/s
Since, it is launched at angle 20 degrees below horizontal,
The initial horizontal velocity
Vx= Vcos(20)=8cos(20)=7.52 m/s
The initial vertical velocity
Vy= -Vsin(20)=-2.74 m/s
In the horizontal direction velocity constant,
Horizontal distance=(7.52)×(3)=22.55 m
Hence ball lands at distance 22.55m from base.
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