Physics problems: Are you smart enough to answers these correctly?

Question

1. A student throws a softball horizontally from a dorm window 15.0 m above the ground. Another student standing 10.0 m away from the dorm catches the ball at a height of 1.50 m above the ground. WHAT IS THE INITIAL VELOCITY OF THE BALL?

2. A certain person can jump a vertical distance of 0.85 m.

a. What is the total time the person is off the ground?

b. With what velocity does the person hit the ground?

3. Three identical balls are thrown or projected from a cliff of a particular height (h) above the horizontal plain. One ball is projected at an angle of 45 degrees to the horizontal, the second ball is thrown horizontally, and the third is thrown at an angle of 30 degrees below the horizontal (that is downward). If all the balls have the same initial speed, which one will have the greatest speed when it hits the ground? (Justify your answer mathematically).

4.A 70.0 kg person stands on a scale in an elevator.What weight in Newton does the scale read if the elevator is:

a. At rest

b. Moving an upward acceleration of 0.525 m/s2(2 is squared)

c. Moving with a downward acceleration of 0.525 m/s2?

5. The coefficient of static and kinetic friction between a floor and 25 kg crate are 0.68 and 0.34, respectively.

If forces of (a) 150 N and (b) 175 N are applied horizontally,

What is the coefficient of static friction between the box and the floor?

Answer 1

1---5*6^1/2(Square root)*m/7*s)

2---a)17^1/2(Square root)m/s

     b)17^1/2*(1+2^1/2)m/s

3---D ball thrown upwards with an angle of 45 degrees from d horizontal will have greatest velocity . Let d ball thrown upwards with an angle of 45 degrees from d horizontal  be ball a1,ball horizontally thrown be a2 & the ball thrown downwards with an angle of 30 degrees with d horizontal be a3. When v say that we want d velocity by which it hits d ground, we mean the vertical component of v(Final Velocity).Let initial velocity be u.Here we take the +ve Y axis(vertical direction) upwards.Thus the value of initial velocity downwards will be -1*(initial velocity in +ve y direction) .For ball A1(Components in d -ve Y direction)---:

u1=-1*u*sin 45= -usin45=-(u*2^1/2)/2

acceleration=g=9.8m/s^2

distance=s=h

v1={(u^2)/2     +   19.6h}^1/2                          ----- (i)

For a2---

u2=u*sin0=0  m/s

acc.=g=9.8  m/s^2

Distance=s=h   m

v2=(19.6h)^1/2    m/s                          ---(ii)

For a3---

u3=u*sin30=u/2   m/s

Distance=s=h  m

Acceleration=g=9.8  m/s^2

v3={(u^2)/4   +   19.6h}^1/2   m/s                    ---(iii)

From (i),(ii),(iii)----v1>v3>v2

Thus , proved dat d ball thrown upwards with an angle of 45 degrees from d horizontal will have greatest velocity.

4---a)686 N

     b)722.89 N

     c)649.25 N

Answer 2

1) Let initial velocity = u,

t = time from when the ball is thrown to when it is caught,

x = horizontal displacement of ball = 10 m,

y = vertical displacement of ball = 15 - 1.5 = 13.5 m

u is horizontal.

Therefore, x = ut ----------(1)

y = (1/2) * g * t^2 --------(2)

Taking square on both sides of (1),

x^2 = u^2 * t^2--------(3)

Divide (3) by (2)

x^2/y = u^2/(g/2)

x^2/y = 2*u^2/g

Or, 2*u^2/g = x^2/y

Or, u^2 = g*x^2/(2 * y)

= 9.8 * 10^2/(2*13.5)

= 36.3

Or, u = 6 m/s (approximately)

2) a)h = 1/2 * gt^2

Or, t = sqrt(2h/g)

h = 0.85 m, g = 9.8 m/s^2

Calculate t

b) When the person is at the top, then his velocity = 0. After covering 0.85 m, he hits the ground. Suppose his velocity now it v.

v^2 = 0 + 2gh

Or, v = sqrt(2gh)

Calculate v

Another method: v = 0 + gt = gt

You have already calculated t. So, calculate v.

I have to go out. So, will answer the rest tomorrow.

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