physics tem 14 tem 14 Part A Arock is tossed straigh up) from the gound with a s

Question

physics tem 14 tem 14 Part A Arock is tossed straigh up) from the gound with a speed of 24 m/s. When it retuns, it fails into a hole 10 n deep You may wand to review (D pages 5154) Whal is the rocks velocihy as it hits the bottiom of the hole? Express your answer with the appropriate units. For help with math skills you may want to review For general problem-solving tips and strategies for this topic you may want to view a Video Tutor Solution of Iime in the ir for a tossed ball Fan ncorrect: One attempt remaining: Try Again Part B Hovi liang is the tock in the air from the Instant t is leased unt t bite the bottom of the hole? Express your an swer with the apptopriate units.

Explanation / Answer

Part (A) : the rock's velocity as it hits the bottom of the hole which is given as ::

using equation of motion 2,

s = ut + 1/2 gt2                         { eq.1 }

where, u = initial velocity = 24 m/s

g = acceleration due to gravity = -9.8 m/s2

s = maximum height reached downward = -10 m

inserting all these values in eq.1,

(-10 m) = (24 m/s) t + (0.5) (-9.8 m/s2) t2

-10 = 24 t - 4.9 t2

rearrnaging above eq.

4.9 t2 - 24 t - 10 = 0                             { eq.2 }

it's an quadratic equation, To find the value of 't' -

t = 5.28 sec    or    t = - 0.38 sec

neglecting the negative value of time, t.

Now, using equation of motion 1 -

v = u + gt                                         { eq.3 }

inserting the values in eq.3,

v = (24 m/s) + (-9.8 m/s2) (5.28 s)

v = (24 m/s) - (51.74 m/s)

v = - 27.7 m/s

Part (B) : Time taken by the rock from instant, it is released until it hits the bottom of the hole which will be given as ::

from equation of motion 2,

t = 5.28 sec

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