We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
37
1
avatar

A ball is thrown horizontally from the top of a building 36.7 m high. The ball strikes the ground at a point 91.9 m from the base of the building. The acceleration of gravity is 9.8 m/s 2. Find the x component of its velocity just before it strikes the ground.

 Oct 29, 2019
 #1
avatar+104932 
+1

We can find the time that it takes the ball to hit the ground  as follows :

 

-36.7  = (1/2) * -(9.8) * t^2

 

-36.7  = -4.9 * t^2

 

(36.7 /4.9)   = t^2    take the positive root

 

t ≈  2.74 sec

 

The x component of the velocity remains constant an can be found as

 

x  = vx * t       

 

Where x is the horizontal displacement, vx  is the horizontal component of the velocity and t is the time it takes the ball to hit the ground....so....

 

91.9  = vx  * 2.74

 

91,9 / 2.74  = vx  ≈  33.6 m / s

 

 

cool cool cool

 Oct 29, 2019

26 Online Users