|Projectile Motion (Science)|

Here are my results:

The results are the same as Melody's but also include the answer to part (c).

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Mmm the horizonal speed will stay 25m/s

$$\\\ddot y=-9.8\\ \ddot y=-9.8t+0\\ y=-4.9t^2+5.5 \\\\ \ddot x=0\\ \dot x=25\\ x=25t$$

The ball will hit the ground when y=0

$$\\-4.9t^2=-5.5\\ t^2=5.5\div 4.9\\ t=\sqrt{5.5\div 4.9}\\$$

$${\sqrt{{\frac{{\mathtt{5.5}}}{{\mathtt{4.9}}}}}} = {\mathtt{1.059\: \!456\: \!926\: \!727\: \!951\: \!9}}$$

1.059 seconds  approx

Horizontal distance = 25t =  $${\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{1.059\: \!456\: \!9}} = {\mathtt{26.486\: \!422\: \!5}}$$       about 26.5 metres

Angle

x velocity = 25m/sec              

 y velocity = $${\mathtt{\,-\,}}\left({\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{1.059\: \!456\: \!9}}\right) = -{\mathtt{10.382\: \!677\: \!62}}$$   = about 10.4m/s

$$tan\theta = 10.4/25$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{10.4}}}{{\mathtt{25}}}}\right)} = {\mathtt{22.587\: \!310\: \!531\: \!366^{\circ}}}$$

angle is about 22.6 degrees 

Thanks Alan, I forgot to include that bit