A ball is thrown horizontally from a height of 5.50 m with an initial speed of 25.0 m/s. (a) How long will it take the ball to reach the ground? (b) At what horizontal distance from the point of release will it strike the ground? (c) What will be the magnitude of its velocity when it strikes the ground? (d) At what direction will it strike the ground?

Maath
May 28, 2015

#1**+10 **

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

Melody
May 28, 2015