Answer: t = 0.25 s
Explanation:
To find the time it will take the ball to travel a vertical height of 5m, we would apply one of Newton's equation of motion shown below
h = ut + 1/2gt^2
where
h is the height or vertical displacement
u is the initial velocity
g is the acceleration due to gravity
t is the time taken to reach height h
From the information given,
h = 5
g = - 9.8 m/s^2 because the ball is decelerating while moving upwards.
Given that the initial velocity is 30m/s and it was launched at an angle of 45 degrees, we would find the y or vertical component of the velocity. Thus,
u = 30sin45
By substituting these values into the equation, we have
5 = 30sin45t - 1/2 x 9.8 x t^2
5 = 21.21t - 4.9t^2
4.9t^2 - 21.21t + 5 = 0
This is a quadratic equation. The standard form of a quadratic equation is expressed as
ax^2 + bx + c = 0
By comparing both equations,
a = 4.9
b = - 21.21
c = 5
We would solve the equation by using the quadratic formula which is expressed as
[tex]\begin{gathered} x\text{ = }\frac{-\text{ b }\pm\sqrt{b^2-4ac}}{2a} \\ x\text{ = }\frac{-\text{ - 21.21 }\pm\sqrt{-\text{ 21.21}^2-4(4.9\text{ }\times5}}{2\times4.9} \\ x\text{ = }\frac{21.21\pm\sqrt{449.8641\text{ - 98}}}{9.8} \\ x\text{ = }\frac{21.21\text{ }\pm18.758}{9.8} \\ x\text{ = }\frac{21.21\text{ + 18.758}}{9.8}\text{ or x = }\frac{21.21\text{ - 18.758}}{9.8} \\ x\text{ = 4.08 or x = 0.25} \end{gathered}[/tex]Replacing x with t, we have
t = 0.25 s