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You lay on the ground shooting Velcro b***s at a velcro wall 30m away. the b***s leave the gun at the inital speed of 23m/s and you are aiming at a 35º angle. How high above the ground will the b***s stick to the wall?

 Jul 7, 2014

Best Answer 

 #1
avatar+118703 
+8

Initially       t=0

¨x=0˙x=23cos350x=0¨y=9.8˙y=23sin350y=0

------------------------------------------------------------------

Ongoing

¨x=0˙x=23cos350x=(23cos350)t     ¨y=9.8˙y=9.8t+23sin350y=9.8t22+(23sin350)ty=4.9t2+(23sin350)t

 

Find y when x=30

When x=30

30=(23cos35)t

t=30/(23cos35)

t=1.592314681

4.9×1.5923146812+23×sin360(35)×1.592314681=8.5824425340714199

The ball will hit the wall 8.58 metres above the floor.

 Jul 7, 2014
 #1
avatar+118703 
+8
Best Answer

Initially       t=0

¨x=0˙x=23cos350x=0¨y=9.8˙y=23sin350y=0

------------------------------------------------------------------

Ongoing

¨x=0˙x=23cos350x=(23cos350)t     ¨y=9.8˙y=9.8t+23sin350y=9.8t22+(23sin350)ty=4.9t2+(23sin350)t

 

Find y when x=30

When x=30

30=(23cos35)t

t=30/(23cos35)

t=1.592314681

4.9×1.5923146812+23×sin360(35)×1.592314681=8.5824425340714199

The ball will hit the wall 8.58 metres above the floor.

Melody Jul 7, 2014
 #2
avatar+33654 
+5

Here's an approach using the constant acceleration kinetics equations:

velcro b***s

 Jul 7, 2014

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