Question

1. How far away will a ball land when thrown from a hill 70 m high at a 45 angle from the horizontal at an initial velocity of 9 m/s?

2. How long will it take for the ball in problem 1 to fall to the ground? Remember the pull of gravity is (-9.8 m/s)

3. What is the maximum height reached by the object?

Answer 1

`28.53\ \text{m}`

`4.48\ \text{s}`

`72.1\ \text{m}`

Step-by-step explanation:

`y_0` = Height from where the ball is thrown =
`70\ \text{m}`

`\theta` = Angle of inclination =
`45^(\circ)`

`u` = Initial velocity =
`9\ \text{m/s}`

`g` = Acceleration due to gravity =
`9.8\ \text{m/s}^2`

Range of projectile is given by

`R=(u\cos \theta )/(g)\left(u\sin \theta +{\sqrt{u^(2)\sin ^(2)\theta +2gy_(0)}}\right)\\\Rightarrow R=(9\cos45^(\circ))/(9.8)\left(9\sin45^(\circ)+\sqrt{9^2\left(\sin45^(\circ)\right)^2+2* 9.8* 70}\right)\\\Rightarrow R=28.53\ \text{m}`

The ball will land
`28.53\ \text{m}` away.

Time of flight is given by

`t=\frac{u\sin \theta +{\sqrt{(u\sin \theta )^(2)+2gy_(0)}}}{g}\\\Rightarrow t=\frac{9\sin 45^(\circ) +{\sqrt{(9\sin 45^(\circ) )^(2)+2* 9.8* 70}}}{9.8}\\\Rightarrow t=4.48\ \text{s}`

The time the ball will stay in the air is
`4.48\ \text{s}`

Maximum height is given by

`h=y_0+(u^2\sin^2\theta)/(2g)\\\Rightarrow h=70+(9^2* \sin^245^(\circ))/(2* 9.8)\\\Rightarrow h=72.1\ \text{m}`

Maximum height the object will reach above the ground is
`72.1\ \text{m}`.