A bowling ball travelling with constant speed hits the pins at the end of a bowl

Question

A bowling ball travelling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound of the ball hitting the pins 2.8 seconds after the ball is released from his hands. What is the speed of the ball assuming the speed of sound is 340 m/s. Show that the time required for a projectile to reach its highest point is the time required for it to return to its original height if air resistance is ignored. You can assume that the projectile is thrown vertically upward. A hypothetical planet has mass 2.8 times that of earth but the same radius. What is the acceleration due to gravity, g near the surface? An aluminum weight is placed on top of a l kg block of wood. The wood has a density of 600 kg/m^3 and is floating in water. What is the mass of aluminum if the top of the wooden block is exactly at the water's surface? A mass m = 5 kg is placed on the end of a spring with spring constant k = 300 N/m a) What is the period of oscillation? If an extra mass of 10 kg is placed on top of the mass already present find b) new period of oscillation. You drop a ball from a height of 2 m and it bounces back to a height of 1.6 m. a) How much energy is lost in the process of bouncing.

Explanation / Answer

1) Time taken by sound to reach the person = 16.5/340 = 0.0485 s

So time taken by ball to reach = 2.8- 0.0485 = 2.75 s

so the speed of the ball = 16.5/2.75= 5.997 m/s

2) Lets say its thrown up with a speed of u , at the highest point speed becomes zero

Vf^2=Vi^2+ 2as

0=u^2-2gs

s=u^2/2g , and Vf=vi+at

0=u-gt

t=u/g

While coming it covers same distance but the starting speed becomes zero

s= vit+ 0.5 at^2

u^2/2g= 0.5 gt^2

t^2= u^2/g^2

t=u/g

So its the same

3) g= GM/R^2 = 2.8*9.81= 27.47 m/s^2

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