Question

A batter hits a ball at 35 degrees above the horizontal and it is caught 4 seconds later 100 meters from home plate. What is the initial velocity (vector) of the ball?

Answer 1

Answer:

$u=(26.5 i + 18.5j) m/s$

Explanation:

The range of a projectile is given by the formula

$d=\frac{u^2}{g} sin 2\theta$

where in this case, we have

d = 100 m is the range

u is the initial speed (the magnitude of the initial velocity)

g = 9.8 m/s^2 is the acceleration of gravity

$\theta = 35^{\circ}$ is the angle of projection

Solving for u, we find:

$u=\sqrt{\frac{dg}{sin 2\theta}}=\sqrt{\frac{(100)(9.8)}{sin(2\cdot 35^{\circ})}}=32.3 m/s$

Now we can easily find the components of the initial velocity:

$u_x = u cos \theta = (32.3)(cos 35^{\circ})=26.5 m/s\\u_y = u sin \theta = (32.3)(sin 35^{\circ})=18.5 m/s$

So, the initial velocity of the ball is

$u=(26.5 i + 18.5  j) m/s$

where i and j are the unit vector indicating the horizontal and vertical direction.