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vredina [299]
3 years ago
7

A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it

reaches home plate, which is a horizontal distance of 60.5 ft away? (Express your answer in units of feet. Neglect air resistance.)
Physics
1 answer:
Lisa [10]3 years ago
3 0

Answer:

Vertical distance=  3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time=  0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

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A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in
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Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

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Now,

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= 10+1.50

= 11.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v = \frac{d}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{11.5}

      =26.08 \ s

Throughout the opposite direction, when the boat seems to be travelling then,

= 10-1.50

= 8.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v=\frac{v}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{8.5}

      =35.29 \ s

hence,

The time taken by the boat will be:

= 26.08+35.29

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