Question

a) A girl threw a ball due west against the wall which is 50.4cm from her position, the
ball bounces due east for 20m.caculate the displacement and distance was covered of the
ball.
b. A ball is thrown vertically up into the air by a boy. After 4 seconds it reaches the
highest point of its path. How fast does the ball leave the boy’s hand

Answer 1

a) who said the ball ended up where it started? It clearly did not.
distance: 50.4cm + 20m
displacement: 20m - 50.4cm

Answer 2

a)

Displacement = 0 (since the ball ends up back where it started)

Distance covered = distance to the wall + distance back from the wall

Distance to the wall = 50.4 cm

Distance back from the wall = 20 m = 2000 cm

Total distance covered = 50.4 + 2000 = 2050.4 cm

b)

Using the formula for vertical displacement:

Δy = vi(t) + (1/2)at^2

At the highest point, the velocity is 0, so:

0 = vi(4) + (1/2)(-9.8)(4)^2

-78.4 = 4vi

vi = -19.6 m/s (negative because it is moving upward)

Therefore, the ball leaves the boy's hand with a speed of 19.6 m/s.

Answer 3

You are correct, I apologize for the error in my previous response. The ball did not end up where it started, so the displacement cannot be zero.

Distance covered = distance to the wall + distance back from the wall

Distance to the wall = 50.4 cm

Distance back from the wall = 20 m

Total distance covered = 50.4 cm + 20 m = 2050.4 cm

Displacement = distance east - distance west

Distance east = 20 m = 2000 cm

Distance west = 50.4 cm

Displacement = 2000 cm - 50.4 cm = 1949.6 cm

Therefore, the displacement of the ball is 1949.6 cm (to the east) and the distance covered is 2050.4 cm.