A ball is thrown vertically upward a velocity of 14 m/s. Calculate 1. The maximum height attained 2.The time taken by it before it reaches the ground again?

Question

Answer

Answer

1.U= 14 m/s
v= 0 m/s (since we need to find maximum height)
a= -9.8 m/s^2

s= (v^2 - u^2)/2a
= (0-196)/-19.6 = -196/19.6 = 10m
Thus the maximum height reached by the stone is 10m.
2.t=?
a=v-u/t,
-9.8=0-14/t,
t=-14/-9.8=1.4285 sec

Answer

Calculation of Maximum Height Attained and Time Taken by a Ball

Given Data:

Initial velocity, u = 14 m/s (upward)

Acceleration due to gravity, g = 9.8 m/s² (downward)

Calculation of Maximum Height:

At maximum height, the final velocity of the ball is zero. Using the kinematic equation:

v² = u² + 2as

where,

v = final velocity = 0

u = initial velocity = 14 m/s (upward)

a = acceleration due to gravity = -9.8 m/s² (downward)

s = maximum height attained

Substituting the given values, we get:

0² = 14² + 2(-9.8)s

Solving for s, we get:

s = 9.8 m

Answer:

The maximum height attained by the ball is 9.8 meters.

Calculation of Time Taken:

Using the kinematic equation:

s = ut + ½at²

where,

s = maximum height attained = 9.8 m

u = initial velocity = 14 m/s (upward)

a = acceleration due to gravity = -9.8 m/s² (downward)

t = time taken to reach maximum height and to fall back to the ground

Substituting the given values, we get:

9.8 = 14t + ½(-9.8)t²

Simplifying the equation, we get:

-4.9t² + 14t - 9.8 = 0

Using the quadratic formula, we get:

t = (-b ± √(b² - 4ac)) / 2a

where,

a = -4.9

b = 14

c = -9.8

Substituting the values, we get:

t = 1.43 s (approx.) or t = 1.02 s (approx.)

Since the time taken cannot be negative, the time taken by the ball before it reaches the ground again is 1.43 seconds (approx.).

Answer:

The time taken by the ball before it reaches the ground again is 1.43 seconds (approx.).

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