Answer:
The time is [tex]t_t = 3.7583 \ s [/tex]
Explanation:
From the question we are told that
The angle is [tex]\theta = 30^o[/tex]
The horizontal distance is [tex]d = 120 \ ft[/tex]
Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as
[tex]v = u + at[/tex]
here a = -g the negative sign is because the direction of motion is against gravity
So
[tex]v = v_i + at[/tex]
Here[tex] v_i [/tex] is the vertical component of the initial velocity of the ball which is mathematically
represented as
[tex]v_i = usin(\theta )[/tex]
So
=> [tex]0 = usin(\theta ) -9.8t[/tex]
Generally the total time taken to travel the 120 ft is mathematically represented as
[tex]t_t = \frac{120}{v_h}[/tex]
Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as
[tex]v_h = u cos(\theta )[/tex]
So
[tex]t_t = \frac{120}{ u cos(\theta )}[/tex]
Generally the time taken to reach the maximum height is
[tex]t = \frac{t_t}{2}[/tex]
=> [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]
=> [tex]t = \frac{60}{ u cos(\theta )} [/tex]
So
[tex]0 = usin(\theta ) -9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) = 9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) * u cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(\theta ) cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(30 ) cos(30) =60* 9.8 [/tex]
[tex] u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2} =588.6 [/tex]
[tex] u^2 *\sqrt{3} =2354.4 [/tex]
[tex] u^2 = 1359.31 [/tex]
[tex] u = 36.87 \ ft/s [/tex]
Substituting this value into the equation for total time
[tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]
[tex]t_t = 3.7583 \ s [/tex]
