# Projectiles question

• May 26th 2011, 09:07 AM
silvercats
Projectiles question
Object is projected from a velocity of 18 m/s and angle of 60 degrees .
take g=10

what is the flying time , Horizontal distance and the maximum height of the object ?
this is what I did which gave the wrong answers
v=u+at
0=v sin Θ-gt/2
t=2v*sin Θ/g
t=2*18 sin 60/10
t=1.8*√3/2
t=1.56

-->
s=18*cos 60*1.56 =14m

(up)
s=18*√3/2 - 5(1.56)²
=3.42m

thanks

book says 0.77s << I think time is wrong in the book:
Horizontal distance 28.02m (i got 14)
maximum height 12.15m (i got 3.42)
• May 26th 2011, 02:07 PM
Soroban
Hello, silvercats!

Quote:

Object is projected with a velocity of 18 m/s and angle of 60 degrees. .Take g=10

What is the (a) flying time, (b) horizontal distance
and (c) the maximum height of the object?

The equations for projectiles are: . $\begin{Bmatrix}x \;=\; (v_o\cos\theta)t \\ y \;=\; h_o + (v_o\sin\theta)t - \frac{1}{2}gt^2 \end{Bmatrix}$

. . where: . $\begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \\ \theta &=& \text{angle of elevation} \end{Bmatrix}$

We have: . $h_o = 0,\;\;v_o = 18,\;\;\theta = 60^o,\;\;g=10$

Hence: . $x \;=\;(18\cos60^o)t \quad\Rightarrow\quad x \;=\;9t$
. . . . . . $y \;=\;(18\sin60^o)t - \tfrac{1}{2}(10)t^2 \quad\Rightarrow\quad y \;=\;9\sqrt{3}t - 5t^2$

(a) Flying time: When does the object strike the ground?
. . .When $y = 0.$

. . $9\sqrt{3}t - 5t^2 \:=\:0 \quad\Rightarrow\quad t(9\sqrt{3} - 5t) \:=\:0 \quad\Rightarrow\quad t \:=\:0,\;\tfrac{9\sqrt{3}}{5}$

. . $\text{The object flies for }\frac{9\sqrt{3}}{5}\text{ seconds.}$

(b) Horizontal distance.
. . . In $\tfrac{9\sqrt{3}}{5}\text{ seconds, the object travels}$
. . . . . $x \:=\:9\left(\tfrac{9\sqrt{3}}{5}\right) \:=\:\frac{81\sqrt{3}}{5}\text{ meters horizontally.}$

(c) Maximum height
. . .The object reaches maximum height in $\tfrac{9\sqrt{3}}{10}\text{ seconds.}$
. . .(Half the flying time.)

$\text{The maximum height is:}$
. . $y \;=\;9\sqrt{3}\left(\tfrac{9\sqrt{3}}{10}\right) - 5\left(\tfrac{9\sqrt{3}}{10}\right)^2 \;=\;\frac{243}{20} \:=\:12.15\text{ meters.}$

• May 26th 2011, 10:09 PM
silvercats
Quote:

Originally Posted by Soroban
(a) Flying time: When does the object strike the ground?
. . .When $y = 0.$

. . $9\sqrt{3}t - 5t^2 \:=\:0 \quad\Rightarrow\quad t(9\sqrt{3} - 5t) \:=\:0 \quad\Rightarrow\quad t \:=\:0,\;\tfrac{9\sqrt{3}}{5}$

. . $\text{The object flies for }\frac{9\sqrt{3}}{5}\text{ seconds.}$

But can't we take like this? (Wondering)
9√3-5t²=0
5t²=9√3
t²=(9√3)/5
t= ± sqrt{(9√3)/5} ?