Question

A batter hits a ball and it is caught 4 seconds later 100 m from home plate. What is the initial velocity (vector!!!) of the ball?(Note: the ball is hit and caught at the same distance from the ground.

Answer 1

Answer:

The initial velocity vector of the ball is;

$\underset{u}{\rightarrow}$ = 25·i + 19.6·j

Explanation:

The given parameters are;

The time of flight of the ball = 4 seconds

The horizontal distance from the plate at which the ball is caught = 100 m

Let 'u', represent the initial velocity, we have;

u × cos(θ) × t = u × cos(θ) × 4 s = 100 m

u × cos(θ) = 25 m/s...(1)

$t = \dfrac{2 \cdot u \cdot sin(\theta)}{g} = \dfrac{2 \cdot u \cdot sin(\theta)}{9.8 \ m/s^2} = 4 \ seconds$

∴ 2·u·sin(θ) = 9.8 m/s² × 4 s = 39.2 m/s

$\therefore \dfrac{2 \cdot u \cdot sin(\theta)}{u \cdot cos(\theta)} = 2 \cdot tan(\theta) = \dfrac{39.2 \ m/s}{25 \ m/s} = 1.568$

tan(θ) = 1.568/2 = 0.784

θ = arctan(0.784) ≈ 38.096°

The direction of the velocity  vector of the ball, θ ≈ 38.096°

From equation (1), we have;

u × cos(θ) = 25 m/s

∴ u = 25 m/s/cos(θ) = 25 m/s/cos(arctan(0.784)) = 31.7672787629 m/s

The magnitude of the initial velocity vector, u = 31.7672787629 m/s

The vertical component of the initial velocity, $u_y$ = u × sin(θ)

∴ $u_y$ = 31.7672787629 × sin(arctan(0.784))

Therefore, the initial velocity vector of the ball is approximately, v = 31.767 m/s in a direction 38.096° above the horizontal, from which we have;

u = uₓ·i + $u_y$·j = 25·i + 19.6·j

$\underset{u}{\rightarrow}$ = 25·i + 19.6·j