Question

A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is made with an initial speed of 19.0 m/s at an angle of 33.0 ∘above the horizontal.

Part a) What is the horizontal component of the ball's velocity just before it is caught?

Part b) How long is the ball in the air?

Answer 1

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the ball = V₀ = 19 m/s

Angle of throw = = 33^o

Initial horizontal velocity of the ball = V_x0

V_x0 = V₀Cos

V_x0 = (19)Cos(33)

V_x0 = 15.935 m/s

There is no force acting on the ball in the horizontal direction hence the velocity in the horizontal direction does not change.

Horizontal velocity just before the ball is caught = V_x1

V_x1 = V_x0

V_x1 = 15.935 m/s

Initial vertical velocity of the ball = V_y0

V_y0 = V₀Sin

V_y0 = (19)Sin(33)

V_y0 = 10.348 m/s

Time of flight of the ball = T

The displacement of the ball in the vertical direction is zero as it starts and ends at the same level.

For the vertical direction,

0 = V_y0T + gT²/2

V_y0 + gT/2 = 0

10.348 + (-9.81)T/2 = 0

T = 2.11 sec

a) Horizontal component of the ball's velocity just before it is caught = 15.935 m/s

b) Time the ball was in air for = 2.11 sec