Question

A catapult launches a large stone from ground level at a speed of 45.2 m/s at an angle of 57.0° with the horizontal. The stone returns to ground level shortly thereafter.
(a) How long is it in the air
s
(b) What maximum height does the stone reach? (Neglect air friction.)
Im

Answer 1

Answer:

a

 $T  =7.7 \ s$

b

   $ s = 73.3 \  m$

Explanation:

From the question we are told that

   The speed is  $u =  45.2  \  m/s$

   The angle is  $\theta  = 57 .0^o$    

   

Generally the velocity of the stone in the y-axis is mathematically evaluated as

        $v_y  = v * sin(\theta )$

=>     $v_y  =  45.2  * sin(57 )$

=>      $v_y  = 37.9 \  m/s$

Now  the time taken for the first flight is mathematically represented as

        $t =  \frac{v_y}{g}$

=>      $t =  \frac{37.9}{9.8}$

=>      $t = 3.87 \  s$

The time taken for the total flight is mathematically represented as

           $T  =  2 * t$

               $T  =  2 * 3.87$

           $T  =7.7 \ s$

From the equation of kinematics we have that  

      $v^2  =  v_y^2  + 2gs$

At maximum heigth v =  0

      $ =  37.9 ^2  + 2 *( -9.8) * s$ The  negative sign is because it is moving against gravity to attain the maximum height

So

        $ s = 73.3 \  m$

       

Answer 2

The time a projectile is in the air and the maximum height reached is required.

The projectile is in the air for 7.73 s.

The maximum height reached by the projectile is 73.24 m.

u = Initial velocity = 45.2 m/s

$\theta$ = Angle of inclination = $57^{\circ}$

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Time of flight is given by

$t=\dfrac{2u\sin\theta}{g}\\\Rightarrow t=\dfrac{2\times 45.2\sin 57^{\circ}}{9.81}\\\Rightarrow t=7.73\ \text{s}$

Maximum height is given by

$h=\dfrac{u^2\sin^2\theta}{2g}\\\Rightarrow h=\dfrac{45.2^2\sin^257^{\circ}}{2\times 9.81}\\\Rightarrow h=73.24\ \text{m}$

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