Question

A home run is hit in such a way that the baseball just clears a wall 21.0 m high, located 130 m from home plate. The ball is hit at an angle of 35.0° to the horizontal, and air resistance is negligible. Find initial speed of the ball. (Assume the ball is hit at a height of 1.00 m above the ground.)Find the speed of the ball when it reaches the wall.

Answer 1

Since the air resistance is negligible, the x and y components can be calculated separately.

The ball is hit 35° to the horizontal so the relation between the x and y components of the initial velocity is:

tan 35° = vy / vx

vy = vx * tan 35°

vx is constant, but vy changes with time.

?x = 130m

?y = 20m (21m - 1m)

Therefore:

?x = vx*t

?y = vy*t +1/2*g*t²

Plug in known values:

130m = vx*t

vx = 130m / t

Substitute for vy then vx:

20m = vx * tan 35° * t + 1/2 * -9.81m/s² * t²

20m = 130m / t * tan 35° * t + 1/2 * -9.81m/s² * t²

The t cancel out:

20m - 130m*tan 35° = 1/2 * -9.81m/s² * t²

t = ?((20m - 130m*tan 35°) / (1/2 * -9.81m/s²))

t = 3.80532863 s

The ball travels 3.80532863 s before going over the fence

a)

We can use the constant x component of the velocity to calculate the initial velocity.

The relation between the velocity and its x component is:

cos 35° = vx / velocity

vx = dx / t = 130m / 3.80532863 s = 34.1626211 m/s

velocity = vx / cos 35° = 34.1626211 m/s / cos 35° = 41.7048597 m/s

b)

vx is constant at 34.1626211 m/s

vy can be calculated using the equation:

v = v0 + a*t

Initial vy = 41.7048597 m/s * sin 35° = 23.9209248 m/s

Final vy = 23.9209248 m/s + (-9.81m/s²) * 3.80532863 s Final vy = -13.4093491 m/s

The speed when the ball reaches the wall is (using Pythagorean Theorem):

v = ?(vx² + vy²) = ?((34.1626211 m/s)² + (-13.4093491 m/s)²)

v = 36.7000725 m/s