Question

What was the minimum speed of the ball when it left the bat when a baseball player hit a home run over the 7.5 m high right field fence 95 m from home plate? Assume the ball was hit from 1.2 meters above the ground, and its initial path was at a 35° angle to the ground.​

Answer 1

Answer:

When babe ruth hit a home over the7.5-m-high right field fence 95 m from plate,  roughly what we the minimum speed  of the ball when it left the ball

Explanation:

Answer 2

Given:

• x = 95 m, a = 0
• y = 7.5 m, a = 9.8

To Find:

• Minimum speed of the ball.

Solution:

• There are two components formed during the process: x-component and y-component.
• The basic formula used for finding both x and y components are:
• $r_f=r_i+v_it+\frac{1}{2}at^2$  
• For x-component the above formula can be written as:
• 95 = 0+v*cosθ*t+(1/2)(0)$t^2$
• 95 = v*t*cosθ
• t = 95/(v*cosθ)
• Now let us find the y-component:
• 7.5 = 1.2+v*sinθ*t-(1/2)(-9.8)$t^2$  
• 6.3 = v*sinθ*t-4.9$t^2$  
• Substitute the value of t we get,
• $6.3 = v*sin(theta) *(\frac{95}{v*cos(theta)} )-4.9(\frac{95}{v*cos(theta)})^2$  
• $6.3 = 95 tan(theta)-4.9(\frac{95^2}{v^2*cos^2(theta)} )$  
• $(95tan(theta)-6.3)*v^2=4.9(\frac{95^2}{cos^2(theta)})$  
• $v = \sqrt{4.9(\frac{95^2}{cos^2(theta)})*\frac{1}{95tan(theta)-6.3} }$  
• WKT, theta = 35
• $v = \sqrt{4.9(\frac{9025}{0.6561})*\frac{1}{60.2}) }$  
• v = $\sqrt{67402.073*0.016} = \sqrt{1078.43317} = 32.83$ m/s

v = 32.83m/s.