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https://prnt.sc/s50ml4

 

Here is the question:

 

 Apr 24, 2020
edited by Alan  Apr 24, 2020
 #1
avatar+30933 
+2

Use the constant acceleration formulae.

 

(a)   Use s = ut + (1/2)at2  where s = displacement in time t, with initial velocity u and acceleration a.

 

For the vertical direction we have:

 a = -32 ft/s2 , u = 70*sin(30°) ft/s  and s = -3 ft  (taking the height when hit as the zero height).

 

So:   -3 = 70*sin(30°)*t - (1/2)*32*t2    or  16t2 - 35t - 3 = 0  (noting that sin(30°) = 1/2)

Solve this for t (using the positive solution of course!).

 

(b)  For the horizontal direction there is zero acceleration, so:   distance = 70*cos(30°)*t  where t is the value found from part (a).

 Apr 24, 2020
 #2
avatar+12433 
+3

Here is my answer. This is how we do it in physics.

laugh

 Apr 24, 2020
edited by Omi67  Apr 24, 2020

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