Question

A ball thrown horizontally at 25 m/s travels a horizontal distance of 50 m before hitting...

A ball thrown horizontally at 25 m/s travels a horizontal distance of 50 m before hitting the ground.
for what height was the ball thrown?
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Answer #1
Concepts and reason

For a projectile motion of object, the horizontal velocity of the object will always be constant throughout the motion.

The fundamental kinematic equations of motion give the relationships between the time, displacement, velocity and acceleration.

According to kinematic equation for object moving with constant velocity, the time taken for the entire motion is the ratio of the displacement of the object with the constant velocity that keeps throughout the transit.

For object projected horizontally have the vertical component of velocity zero.

The acceleration of a projectile throughout its transit will be equal to the acceleration due to gravity. The acceleration due to gravity always directs towards Earth.

Fundamentals

The horizontal velocity of a projectile is always same throughout the motion. Hence the time, t taken for the transit can be found by,

t=xvt = \frac{x}{v}

Here, x is the horizontal range of the projectile and v is the initial velocity of the projectile.

The vertical distance, y travelled by an object thrown horizontally is,

y=12gt2y = \frac{1}{2}g{t^2}

Here, g is the acceleration due to gravity.

Time taken for completing the horizontal transit

The time, t taken for the ball to travel the horizontal distance is,

t=xvt = \frac{x}{v}

Substitute the horizontal distance as 50 m and the initial velocity as 25 m/s to calculate the time taken for the ball to travel the horizontal distance. The time is,

t=50m25m/st=2s\begin{array}{l}\\t = \frac{{50\,{\rm{m}}}}{{25\,{\rm{m/s}}}}\\\\t = 2\,{\rm{s}}\\\end{array}

The vertical height the ball horizontally thrown from is,

y=12gt2y = \frac{1}{2}g{t^2}

Substitute acceleration due to gravity as 9.81m/s2 and the time taken as 2 s. The vertical height is,

y=12(9.81m/s2)(2s)2y=19.6m\begin{array}{l}\\y = \frac{1}{2}\left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right){\left( {2\,{\rm{s}}} \right)^2}\\\\y = 19.6\,{\rm{m}}\\\end{array}

The ball is thrown from a height of 19.6 m.

Ans:

The ball is thrown from a height of 19.6 m.

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Answer #2

the ball is going 69mph

source: atanervatidanetsineth
answered by: pookalina
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