Asked by 0ooooooooo0
A ball is kicked from ground level so thta is intiial velocityis 10ms-1 AT AN angleof 60 degrees above the horizontal. It hits a wall which is at a distance of 8m from the initial position of the ball
a)show that the ballhits the wall 1.6s after it was kicked
b)find the heightof the ball as it hits the wall
c)find the speed of the ball as it hits the wall.
Please help with a clear explanation i have no idea
a)show that the ballhits the wall 1.6s after it was kicked
b)find the heightof the ball as it hits the wall
c)find the speed of the ball as it hits the wall.
Please help with a clear explanation i have no idea
Answers
Answered by
Steve
the horizontal speed is constant at 5 m/s, so it takes 8/5 = 1.6 seconds to hit the wall
the height
h(t) = 10 sin(pi/3)t - 4.9t^2
plug in t=1.6
the speed is the combination of vertical and horizontal velocities, so
vx = 10 cos(pi/3) = 5
vy = 10 sin(pi/3) - 9.8t
the speed
v^2 = (vx)^2 + (vy)^2
the height
h(t) = 10 sin(pi/3)t - 4.9t^2
plug in t=1.6
the speed is the combination of vertical and horizontal velocities, so
vx = 10 cos(pi/3) = 5
vy = 10 sin(pi/3) - 9.8t
the speed
v^2 = (vx)^2 + (vy)^2
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