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When a rock thrown straight up climbs for 3.00 s before falling. Neglecting air resistance with what velocity did the rock strike the ground?
MadisonBgp5487
You can use Newton's equations of motion:
At the top of the climb its velocity u = 0 m/s
Its acceleration is acceleration due to gravity a ≈ 9.8 m/s
Time of descent t = time of ascent = 3.00 s
(I'll assume positive is towards the ground)
v = u + at
≈ 0 m/s + 9.8 m/s² × 3.00 s
= 29.4 m/s
HOWEVER, this is the velocity (towards the ground) reached when the rock has returned to height from which it was thrown (released) above the ground - unless the rock was "thrown" by an explosive force at ground level, the rock will not have reached the ground at this point: there is still the distance from which it was "thrown".
Which means its final velocity at ground level can be found using:
v² = u² + 2as
v = velocity it hits the ground
u ≈ 29.4 m/s (as found above)
s = distance above ground from which the rock was "thrown" = height_of_throw m
a = acceleration due to gravity ≈ 9.8 m/s
→ v² = u² + 2as
→ v ≈ √((29.4 m/s)² + 19.6 m/s² × height_of_throw m)
= √(864.36 + 19.6 × height_of_throw) m/s
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∙ 6y ago
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