Problem:
Jamilla throws a stone horizontally off a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0m from the edge of the pier when it hits the water, how high above the water's surface was the stone when Jamilla threw it?
Related Equations:
displacement = velocity x time
displacement = (v1) (t) + (a) (t2) /2
t= square root of (2) (d) / a
Solution:
t= square root of (2) (d) /a
t= square root of (2) (40.0m) /9.8m/s2
t= square root of 80m/ 9.8m/s2
t= square root of 8.16
t= 2.9s
d= (v)(t)
d= (20m/s) (2.9s)
d= 58.0m
t= time
d= displacement
a= acceleration
Having trouble with the equations again. I want to know whether the equation I'm using is right or do I have to use a different equation?
Therefore the stone was 58.0m high above the ground when Jamilla threw it.
Jamilla throws a stone horizontally off a pier into the ocean at a velocity of 20 m/s. If the stone is 40.0m from the edge of the pier when it hits the water, how high above the water's surface was the stone when Jamilla threw it?
Related Equations:
displacement = velocity x time
displacement = (v1) (t) + (a) (t2) /2
t= square root of (2) (d) / a
Solution:
t= square root of (2) (d) /a
t= square root of (2) (40.0m) /9.8m/s2
t= square root of 80m/ 9.8m/s2
t= square root of 8.16
t= 2.9s
d= (v)(t)
d= (20m/s) (2.9s)
d= 58.0m
t= time
d= displacement
a= acceleration
Having trouble with the equations again. I want to know whether the equation I'm using is right or do I have to use a different equation?
Therefore the stone was 58.0m high above the ground when Jamilla threw it.
1 commentaires:
The equation is work break it down into horizontal velocity and vertical velocity. find time by using t=d/v (since in horizontal velocity their is 0 acceleration you can use this formula).Use horizontal distance over horizontal velocity. So, t=40/20 which =2s. Then after you have found time switch to vertical motion and find the height/distance. Use d=v1(t)+1/2(a)(t^2), u can not use the classic d=v/t cuz in vertical velocity their is acceleration due to gravity. Also, on vertical motion v1 or initial velocity is always 0. So, 0(2)+1/2(-9.8)(2^2)= -19.6m. therefore Jamila was 19.6m when he threw the stone. Side not: the reason acceleration due to gravity is always -9.8m/s^2 which is why a=-9.8m/s^2.
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