An object was thrown at a certain angle above the ground.It reaches a maximum height of 42.50 meters and hits back the ground 76
meters
1.find the time of flight
2.what is the initial velocity
3.find the angle of the projectile
1.find the time of flight
2.what is the initial velocity
3.find the angle of the projectile
1 answer:
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Answer:
Position = behind the mirror
Nature = Virtual and Erect
Size = : Diminished
Explanation:
Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.
Object distance = u = -20 cm
Focal length = f = Radius of curvature/2 = 30/2 = 15 cm
We have to use mirror formula to find image distance.
Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.
Magnification =
Height of the object = 5 cm
Height of the image =
Since the height of the image is positive and less than the size of object,it is erect and diminished.