A ball is thrown straight up and rises to a maximum height of 22 m. At what heig

Question

A ball is thrown straight up and rises to a maximum height of 22 m. At what height is the speed of the ball equal to half of its initial value? Assume that the ball starts at a height of 1.8 m above the ground.

____m

An airplane has a velocity relative to the air of 180 m/s in a westerly direction. If the wind has a speed relative to the ground of 50 m/s directed to the north, what is the speed of the plane relative to the ground?
_____m/s

Consider the system of blocks in the figure below, with m2 = 4.9 kg and ? = 31

A ball is thrown straight up and rises to a maximum height of 22 m. At what height is the speed of the ball equal to half of its initial value? Assume that the ball starts at a height of 1.8 m above the ground. ____m An airplane has a velocity relative to the air of 180 m/s in a westerly direction. If the wind has a speed relative to the ground of 50 m/s directed to the north, what is the speed of the plane relative to the ground? _____m/s Consider the system of blocks in the figure below, with m2 = 4.9 kg and ? = 31°. If the coefficient of static friction between block #1 and the inclined plane is ?S = 0.24, what is the largest mass m1 for which the blocks will remain at rest? _______ kg

Explanation / Answer

1. Max height attained from initial pt, h1 = 22-1.8 = 20.2 m

u^2/2g = 20.2

u = sqrt(2*9.8*20.2)

= 19.89 m/s

Kinematinc Eqn is

v^2 - u^2 = 2as

(u/2)^2 - u^2 = -2gs

s = 3u^2/8g

= 3*19.89^ / 2*9.8

= 15.15 m

The height above the ground heightwhere velocitiy is half of its initial value

h2 = s+1.8 m

     = 15.15 + 1.8

       = 16.95 m --> Answer

2. Angle between airplane velocity and air velocity = 90
So, its resultant
= sqrt (180^2 + 50^2)
= sqrt ( 34900) = 186.81 m/s
Answer: velocity of plane relative to ground =186.81 m/s --> Answer

3. the forces acting on m2 = F2 = fg (force of gravity)+ fs(force of friction)

HERE fg on an inclined plane = mgsin31 = 4.9*9.8*sin 31 = 24.732
fs =mu(s)*mgcos31 = 0.24*4.9*9.8*cos31 = 9.8787

Hence forces acting on m2 = 24.732+9.8787 = 34.611
force on m1 = F1= m1*g
at equiliberium F1=F2 ,

Therefore F1= 34.611 N
Hence m1= 34.611 / 9.8 = 3.532 Kg --> Answer
m1 cannot be greater that 3.532 kg

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