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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?

 May 28, 2015

Best Answer 

 #2
avatar+27653 
+15

Here are my results:

 projectile motion:

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

.

 May 28, 2015
 #1
avatar+100007 
+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 

 May 28, 2015
 #2
avatar+27653 
+15
Best Answer

Here are my results:

 projectile motion:

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

.

Alan May 28, 2015
 #3
avatar+100007 
0

Thanks Alan, I forgot to include that bit   

 May 29, 2015

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