This time, William Tell is shooting at an apple that hangs on a tree. The apple

Question

This time, William Tell is shooting at an apple that hangs on a tree. The apple is a horizontal distance of 20.0m away and at a height of 4.00m above the ground.

If the arrow is released from a height of 1.00m above the ground and hits the apple 0.510s later, what is the arrow

This time, William Tell is shooting at an apple that hangs on a tree. The apple is a horizontal distance of 20.0m away and at a height of 4.00m above the ground. If the arrow is released from a height of 1.00m above the ground and hits the apple 0.510s later, what is the arrow?s initial velocity?

Explanation / Answer

Let velocity of projection be v m/s at an angle of ? with the horizontal. Then its---

Vertical component = v sin ? and

Horizontal Component = v cos ?

According to the question the arrow covers a horizontal distance of 20 m to hit the apple in 0.5 s

=> ( v cos ? ) * 0.5 = 20

=> v cos ? = 40 ............. (1)

Again it rises to a height of ( 4 - 1 ) = 3 m above the point of projection in 0.5 s.

=> ( v sin ? ) * 0.5 = 3

=> v sin ? = 6 ........... (2)

From (1) and (2) we get ---

=> tan ? = 6/40 = 3/20 = 0.1500

=> ? = 8.53076561 degree

Hence sin ? = 0.148340452 and cos ? = 0.988936352

From (1)

=> v cos ? = 40 ............. (1)

=> v = 40 / cos ? = 40.447 m/s

inclined at an angle of = 8.53076561 degree with the horizontal

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