Question

ball is launched with an initial speed of 30 m/s making an angle of 45° above the horizontal. How long does it take the ball to reach a vertical displacement Δy= +5 m for the first time?*1.12 s0.54s0.9s0.25s0.7 s

Answer 1

Answer: t = 0.25 s

Step-by-step 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

`\begin{gathered} x\text{ = }\frac{-\text{ b }\pm√(b^2-4ac)}{2a} \\ x\text{ = }\frac{-\text{ - 21.21 }\pm\sqrt{-\text{ 21.21}^2-4(4.9\text{ }*5}}{2*4.9} \\ 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}`

Replacing x with t, we have

t = 0.25 s