A bullet moving 150 m/s strikes a brick wall and comes to rest 3.0 cm inside a b

Question

A bullet moving 150 m/s strikes a brick wall and comes to rest 3.0 cm inside a brick. If the acceleration is uniform, then the acceleration is how many g's ? (is what value times g, the acceleration due to gravity?) Let the direction away from the wall be the positive direction. A cannonball is shot vertically upwards with initial speed 20_m/s. Neglecting air resistance, what max height should the ball reach? How long does it take the ball to reach half of this height? Suppose an object in an experiment starts at the origin and has velocity (as a function of time) given by v = 2.36t^3 + 3.461. Assume SI units are being used. Find the position x of the object at t =3.88 s.

Explanation / Answer

1) Correct option is (f)

45 km^2/h^3 = 45(1000)^2 / (3600)^3 m^2/s^3

= 0.0009645 m^2/s^3

2) Correct option is (d)

u = 150 m/s , s =3 cm

from kinematic equaiton v^2 -u^2 =2as

a = (150*150)/(2*0.03) = 375000

a/g =375000/9.8 =38270

3) Correct option is (e)

u =20 m/s , g =9.8 m/s^2

h = u^2/ 2g = (20*20)/(2*9.8)

h = 20.41 m

4) Correct option is (b)

s =ut+(1/2)at^2

(20.41)/2 = 20*t -(4.9)t^2

t =0.5977s

5) correct option is (b)

vx =2.36t^3 +3.46t

integration of vx gives position

x = 2.36t^4/4 + 3.46t^2/2

x = 0.59 t^4 + 1.73 t^2

for t =3.88

x =(0.59)(3.88)^4 +(1.73)(3.88)^2

x= 159.8 m

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