Question

EXAMPLE 3.4 A home-run hit A home-run baseball is hit with an initial speed v0 5 37.0 m/s at an initial angle u0 5 53.1°. (a) Find the ball’s position, and the magnitude and direction of its velocity, when t 5 2.00 s. (b) Find the time when the ball reaches the highest point of its flight, and find its height h at that point. (c) Find the horizontal range R (the horizontal distance from the starting point to the point where the ball hits the ground).

Answer 1

Position 44.44m and speed is 24.36 m/s with at an angle to the horizontal 87.6º

Step-by-step explanation:

In the launch of projectiles the movements in the x and y axes can be analyzed separately, we start by decomposing the inical velocity in its components, using trigonometry

Vox = Vo cos θ

Voy = Vo sin θ

Vox = 37 cos 53.1 = 22.22 m / s

Voy = 37 sin 53.1 = 29.59 m / s

a) We calculate the position and speed for 2.00s

X = Vox t

X = 22.22 2 = 44.44 m

The velocity on the x-axis is constan since there is not acceleration, on the y-axis we have the acceleration of gravity

Vx = Vox

Vy = Voy - g t

Vx = 22.22 m / s

Vy = 29.59 - 9.8 2.00

Vy = 9.99 m / s

V² = Vx² + Vy²

θ = tan-1 (Vy / Vx)

V =√ ( 22,22² + 9,99²)

V = 24.36 m / s

θ = tan-1 (9.99 / 22.22)

θ = 87.6º

b) In the highest point the vertical speed must be zero (Vy = 0 m/s)

Vy = Voy - g t

t = Voy / g

t = 29.59 / 9.8

t = 3.02 s

To calculate the height we use the height equation with the time found

Y = Voy t - ½ g t2

Ymax = 29.59 3.02 - ½ 9.8 3.02²

Ymax = 44.67 m

c) In this part the equation of the launch range is used

R = Vo2 sin 2θ / g

R = 372 sin (2 53.1) /9.8

R = 134.15 m