Question

A projectile is launched with an initial velocity of 25 m/s at an angle of 30° above the horizontal. The projectile reaches maximum height at point p then falls to point a, which is 65 m below the cliff. What is the horizontal distance that it travels (labeled x in the diagram)

Answer 1

Answer:

x ≈ 56 m

Explanation:

vertical initial velocity =$v_{0y}(t)$ = 25 m/s* sin(30°)= 12.5 m/s

height = h

$h =v_{0y}t+\frac{at^{2}}{2} \\\\65m = 12.5m/s*t + \frac{9.8m/s^{2}*t^{2}}{2} \\\\t=2.584 s$

t- time is found solving quadratic equation.

horizontal velocity = $v_{0x}=25m/s*cos(30^{o})=21.65 m/s$

Horizontal velocity is constant, so distance $x=v_{0x}*t =21.65 m/s *2.584 s=55.9 = 56 m$

Answer 2

The horizontal distance travelled by the object will be "111.07 m".

According to the question,

$u_v = u sin\Theta$

    $= 25 \ sin30^{\circ}$

    $= 12.5 \ m/s$

As we know,

→  $s = ut +\frac{1}{2} at^2$

 $-y = u_vt-\frac{1}{2}gt^2$

$-65 =12.5t - \frac{1}{2}\times 9.8t^2$

   $0 = 4.9t^2-12.5t-65$

    $t = \frac{12.5 \pm \sqrt{(-12.5)^2-4\times 4.9\times (-65)} }{2\times 4.9}$

       $= 5.13 \ s$

hence,

The horizontal distance will be:

→ $u_x = ucos \Theta$

       $= 25 cos 30^{\circ}$

       $= 21.65 \ m/s$

or,

→ $x = 21.65\times 5.13$

     $= 111.07 \ m$

Thus the response above is right.  

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