Answer:
4.25m
Explanation:
The horizontal and vertical component of the velocity as it leaves the roof is
[tex]v_v = vsin30^0 = 4*0.5 = 2 m/s[/tex]
[tex]v_h = vcos30^0 = 4*0.866 = 3.64 m/s[/tex]
Neglect air resistance, then gravitational acceleration g = 10m/s2 is the only thing that affect the vertical motion of the hammer. We can use the following equation of motion to solve for the time t of dropping
[tex]h = v_vt + gt^2/2[/tex]
where h = 10m is the distance of dropping before the hammer reaches the ground
[tex]10 = 2t + 5t^2[/tex]
[tex]5t^2 + 2t - 10 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{-2\pm \sqrt{(2)^2 - 4*(5)*(-10)}}{2*(5)}[/tex]
[tex]t= \frac{-2\pm14.28}{10}[/tex]
t = 1.23 or t = -1.63
Since t can only be positive we will pick t = 1.23
t is also the time that it takes to travel horizontally at the constant rate of 3.64 m/s
So the horizontal distance traveled by the hammer is
3.64*1.23 = 4.25 m
The Horizontal distance travelled by the hammer is : 4.26 m
Given data :
Speed of hammer = 4 m/s
Angle made by roof with the horizontal = 30°
Minimum distance of roof from ground = 10 m
g = 10 m/s²
Considering the y-direction ( vertical ) to determine the Time ( t )
s = ut + 1/2 at²
- 10 = -4sin30 t + 1/2 * 10t²
10 = 2t + 5t²
5t² + 2t - 10 = 0 ----- ( 1 )
Resolve equation ( 1 )
therefore :
t = 1.228 sec
The Horizontal distance is calculated using this formula
4 (cos 30°) t = 4 * cos 30 * 1.228
= 4.26 m
Hence we can conclude that the Horizontal distance travelled by the hammer is : 4.26 m
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