A projectile is shot from the ground at an angle of $\frac{π}{4}$ $pi/4$ and a speed of $6 \frac{m}{s}$ $6 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Question

A projectile is shot from the ground at an angle of $\frac{π}{4}$ $pi/4$ and a speed of $6 \frac{m}{s}$ $6 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer

Impact of this question

1624 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-17T12:22:05

$2.6 m$ $2.6m$

Explanation:

Split up the velocity into its horizontal and vertical components using trigonometry,

$V_{h} = 6 \frac{m}{s} \cdot cos (\frac{π}{4}) = 6 cos 45$ $V_h=6m/s*cos(pi/4)=6cos45$
$V_{v} = 6 \frac{m}{s} \cdot sin (\frac{π}{4}) = 6 sin 45$ $V_v=6m/s*sin(pi/4)=6sin45$

Now you should work out the halfway point, where upwards velocity becomes $0$ $0$ and the projectile reaches its maximum point

$v = u + a t$ $v=u+at$
$0 = V_{v} + gt$ $0=V_v+"gt"$ ,

where the acceleration of gravity $g = - 9.8$ $g=-9.8$

$0 = 6 sin 45 - 9.8 t$ $0=6sin45-9.8t$

$t = \frac{6 sin 45}{9.8}$ $t=(6sin45)/9.8$
$t = 0.433 s$ $t=0.433s$

Now you can work out the horizontal distance by

$s = v t$ $s=vt$
$s = V_{h} \cdot 0.433$ $s=V_h*0.433$
$= 6 cos 45 \cdot 0.433 = 1.84 m$ $=6cos45*0.433=1.84m$

and the vertical distance by

$s = V_{v} \cdot 0.433$ $s=V_v*0.433$
$= 6 sin 45 \cdot 0.433 = 1.84 m$ $=6sin45*0.433=1.84m$ .

The overall distance is found by Pythagoras' theorem,

$c = \sqrt{a^{2} + b^{2}}$ $c=sqrt(a^2+b^2)$

$c = \sqrt{{1.84}^{2} + {1.84}^{2}} = 2.6 m$ $c=sqrt(1.84^2+1.84^2)=2.6m$

Science

Math

Humanities

... and beyond

A projectile is shot from the ground at an angle of $\frac{π}{4}$ $pi/4$ and a speed of $6 \frac{m}{s}$ $6 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A projectile is shot from the ground at an angle of pi/4 π4pi/4 and a speed of 6 m/s6ms6 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer

Explanation:

Related questions

Impact of this question

A projectile is shot from the ground at an angle of $\frac{π}{4}$ $pi/4$ and a speed of $6 \frac{m}{s}$ $6 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?