the first ball. (a) At what speed was the first ball thrown ) At what speed was

Question

the first ball. (a) At what speed was the first ball thrown ) At what speed was the second ball thrown? A home run is hit in such a way that the baseball just lears a wall 21 m high, located 130 m from home plate The ball is hit at an angle of 35° to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (Assume the ball is hit at a height of 1.0 m above the ground.) . A 2.00-m-tall basketball player is standing on the flo or 10.0rm 20 8

Explanation / Answer

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)

Already calculated: t = 3.80532863 s

c)

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

Answers:

a) Initial ball speed: 41.7 m/s

b) Time for ball to reach wall: 3.81 s

c) x velocity-component: 34.2 m/s

y velocity-component: -13.4 m/s

Speed of the ball: 36.7 m/s

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