The ball of a pinball machine is accelerated from rest to 6.2 in/sec in a distan

Question

The ball of a pinball machine is accelerated from rest to 6.2 in/sec in a distance of 3.24 in. Determine the acceleration of the ball A newly developed electric car is claimed to reach 103 km/hr in 2.45 sec. Calculate its average acceleration (m/s^2) if the car starts from rest. A bullet is shot straight in the air with an initial velocity of 185 ft/sec, Calculate how high (absolute in ft) it w go up before it stops and reverses its motion. Assume g 32.2ft/sec^2 and air resistance is negligible. An object with negligible air resistance is dropped from a height of 73 m. Determine the velocity at which it lands. A heavy object with negligible buoyancy enters water at a speed of 10 m/sec. Assuming a constant deceleration rate of -5.2 m/s^2 in water. Calculate how deep does it travel below the surface (absolute value of depth) before its speed is reduced to 7.3 m/s A heavy object with negligible buoyancy is dropped from a height of 114 m above the water level of a lake. Assuming a constant deceleration rate of -28.6 m/s^2 after hitting the water calculate how deep does it travel below the surface (absolute value of depth) before its speed is reduced to 3 84 m/s.

Explanation / Answer

a) Use the equation of motion :

v^2 = u^2 +2as

So, 6.2^2 = 0^2 +2*a*3.24

So, a = 5.93 in/s2 <--------answer

b)

USe the formula : v = u + at

So, (103*1000/3600) = 0 + a*2.45

So, a = 11.7 m/s2

c)

height reached, h = u^2/2g

So, h = 185^2/(2*32.2) = 531.4 ft

d)

v^2 = u^2 + 2as

So, v^2 = 0^2 + 2*9.8*73

So,v = 37.8 m/s

e)

v = u + at

So, 7.3 = 10 - 5.2 *t

So, t = 0.52 s

f)

speed with which it hits water , u = sqrt(2gh)

= sqrt(2*9.8*114)

= 47.3 m/s

Now, use the equation of motion, v^2 = u^2 + 2as

So, 3.84^2 = 47.3^2 - 2*28.6*s

So, s = 38.9 m <-----answer

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