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Ball will hit the ground after a time of 1.296 s
Explanation:
initial velocity of ball= Vi=0
g= 9.8 m/s²
height =h= 27 ft=8.23 m
using the kinematic equation
h= Vi t + 1/2 gt²
8.23=0(t) + 1/2 (9.8)t²
t²=1.6796
t=1.296 s
B) Distance: the farther, the weaker!
Answer:
The required acceleration is
m/s²
Explanation:
Given
To determine
Acceleration a = ?
We know that acceleration is produced when a force is applied to a body.
The acceleration can be determined using the formula

where
now substituting F = 250 , and m = 221 in the formula


switch the equation

Divide both sides by 221

simplify

m/s²
Therefore, the required acceleration is
m/s²
Refer to the diagram shown below.
The following discussion assumes a simplistic analysis that ignores air resistance and variations in the terrain that the missile travels over.
Let the launch velocity be V₀ at an angle of θ relative to the horizontal.
The horizontal component of velocity is V₀ cosθ.
If the time of flight is

, then

where r = the range of the missile.
Also, the time, t, when the missile is at ground level is given by

where g = acceleration due to gravity.
t = 0 corresponds to when the missile is launched. Therefore

Therefore

Typically, θ=45° to achieve maximum range, so that

This analysis is more applicable to a scud missile rather than a powered, guided missile.
Answer:

Usually, θ=45°