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A baseball is batted from a height of 1.09 m with a speed of
39.7 m/s at an initial angle of 17.8˚ above the horizontal.

a) Calculate the horizontal and vertical components of the ball’s initial velocity.
b) What is the maximum height above the ground reached by the ball?
Assume a level baseball field.
c) How far off course would the ball be carried after 2.00 s if there was a 4.0 m/s crosswind?
d) What is the ball’s velocity after 2.00s if here is no crosswind?

Sagot :

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(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Horizontal and vertical components of the ball's velocity

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

Maximum height reached by the ball

[tex]H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m[/tex]

Maximum height above ground = 7.51 + 1.09 = 8.6 m

Distance off course after 2 second

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

[tex]V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s[/tex]

Resultant speed of the ball and crosswind

[tex]V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s[/tex]

Distance off course the ball would be carried

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: https://brainly.com/question/11049671

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