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
a) who said the ball ended up where it started? It clearly did not.
distance: 50.4cm + 20m
displacement: 20m - 50.4cm
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.
a) To calculate the displacement and distance covered by the ball, we can break the motion into two parts: the initial throw and the bounce back.
1. Displacement:
- The girl throws the ball due west toward the wall, which is a displacement of -50.4 cm (negative because it's in the opposite direction).
- The ball then bounces due east for 20 m, which gives a displacement of +20 m (positive because it's in the same direction).
- The total displacement can be calculated by adding the displacements: -50.4 cm + 20 m.
Converting the units to the same system:
- 1 m = 100 cm
- 50.4 cm = 50.4 cm / 100 = 0.504 m
Therefore, the total displacement is -0.504 m + 20 m = 19.496 m (rounded to three decimal places).
2. Distance:
- The distance traveled during the initial throw is the same as the displacement, which is 0.504 m.
- The distance traveled during the bounce back is also 20 m.
- The total distance covered is the sum of these: 0.504 m + 20 m = 20.504 m (rounded to three decimal places).
b) To determine the speed at which the ball leaves the boy's hand, we need to consider the upward motion of the ball during the first 4 seconds.
- The ball reaches its highest point after 4 seconds, so the time of flight is 4 seconds.
- At the highest point of its path, the vertical velocity of the ball is 0 m/s. This is because at the highest point, the ball momentarily stops before starting to fall back down.
- Using the equations of motion under constant acceleration, we know that the initial vertical velocity (u) is equal to the final vertical velocity (v) at the highest point, and the acceleration (a) is equal to -9.8 m/s^2 (acceleration due to gravity).
Using the equation v = u + at, where v = 0 for this case:
0 = u - 9.8 m/s^2 * 4 s
Rearranging the equation to solve for u, we get:
u = 9.8 m/s^2 * 4 s
Simplifying, we find that the initial vertical velocity of the ball is 39.2 m/s.
Therefore, the ball leaves the boy's hand with an initial vertical velocity of 39.2 m/s.
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.