Question

A home run is hit in such a way that the baseball just clears a wall 21 m tall, located at a distance 130 m from the home plate. the ball is hit at a angle of 35 deg to the horizontal and there is no air resistance. Find(a) the initial speed of the ball (b) the time it takes to reach the wall and (c) the velocity components and the speed of the ball when it reaches the wall.( assume the ball was hit at a height of 1 m above the ground. Please show me the steps you take to solve this.

Answer 1

a) v₀ = 41.68 m / s , b) t = 3,808 s

, c) vₓ = 34.14 m / s , v_{y} = -13.41 m / s

Step-by-step explanation:

This is a projectile launch problem

a) To find the initial speed, let's work each component independently

X axis

As there is no friction the speed is constant

v₀ₓ = x / t

Y Axis

v_{y}² = ² - 2 g (y-y₀)

v_{y} = v_{oy} - g t

y = y₀ + v_{oy} t - ½ g t²

Speed ​​components can be found with trigonometry

sin 35 = voy / v₀

cos 35 = v₀ₓ / v₀

v₀ₓ = v₀ cos 35

v_{oy} = v₀ sin 35

At the point where the wall is, time is the same for both movements

t = x / v₀ cos 35

(y-y₀) = v₀ sin35 t - ½ g t²

We replace

y -y₀ = v₀ sin 35 x /v₀ cos 35 - ½ g (x /v₀ cos 35)²

y-yo = x tan 35 - ½ g x² / v₀² cos² 35

Let's clear the initial speed

1 / v₀² = (x tan35 - (y-y₀)) 2 cos² 35 / g x²

v₀² = g x² / 2 (x tan35 - (y-y₀)) cos² 35

Let's calculate

v₀² = 9.8 130² / (2 (130 tan 35 - (21 -1 )) cos² 35

v₀ =√ (165620 / 95.3196)

v₀ = 41.68 m / s

b) Let's use the x component of the movement

t = x / v₀ cos 35

t = 130 / (41.68 cos 35)

t = 3,808 s

c) the velocity on the x axis is constant

vₓ = v₀ cos 35

vₓ = 41.68 cos 35

vₓ = 34.14 m / s

The speed on the y axis is

= v₀ sin35 - g t

v_{y} = 41.68 sin 35 - 9.8 3,808

v_{y} = -13.41 m / s

The negative sign indicates that the ball is descending