# A mis-hit golf ball flies straight up in the air. Exactly 4.0 seconds later it lands right next to the tee. How high up did the golf ball go?

19.6 m

Explanation:

The total motion of the golf ball lasts 4.0 seconds: since the motion is symmetrical, it takes 2.0 s for the ball to reach the highest point and then another 2.0 s to land back on the tee.

Therefore, we can just analyze the second half of the motion that lasts

t = 2.0 s

During this time, the vertical distance covered by the ball is given by the equation:

where

u = 0 is the initial velocity (zero because the ball starts from its highest point, where the velocity is zero)

t = 2.0 s

g = 9.8 m/s^2 is the acceleration of gravity

Solving for d, we find:

So, the ball reaches a maximum height of 19.6 m.

## Related Questions

A small plastic bead has been charged to -15nC. A) What is the magnitude of the acceleration of a proton that is 0.90cm from the center of the bead?

B) In which direction does the acceleration vector point?

C) What is the magnitude of the acceleration of an electron that is 0.90cm from the center of the bead?

D) In which direction does the acceleration vector point?

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

(C) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.

A particular Alnico (aluminum, cobalt, nickel, and iron) bar magnet (magnet A) has a mass of 10 g. It produces a magnetic field of magnitude 3e-05 T at a location 0.17 m from the center of the magnet, on the axis of the magnet. (a) Approximately what is the magnitude of the magnetic field of magnet A a distance of 0.51 m from the center of the magnet, along the same axis

The magnet applies a magnetic field  0.7 * 10^-5 T at distance 0.51 m

Explanation:

Solution

The magnetic field is produced due to the orientation of the electrons in their orbits in the material of the magnet where the bar magnet is a magnetic dipole where the magnetic field produced due to the magnet is given by

B_a=μ_o*2μ/4*π*r^3                                                       (1)

Where μ is the magnetic dipole, the term μ_o/4*π is constant and equals 1 x 10 ^-7 T.m^2/C.m/s  and r is the distance from the magnet to the location where  the magnetic field is applied. μ is the magnetic dipole.

As shown by equation (1), the magnetic field is inversely proportional to

r_3 (B ∝ 1/r^3). therefore, for two instants r1 and r2, we could get the next r3 relationship in the form

B_1/B_2=r_2^2/r_1^3

B_2=(r_1^3/r_2^3)B_1                                                       (2)

Where  B_1 = 3 x 10^ -5 T, r_1 = 0.17 m and r_2 = 0.51 m. Now we can plug our values for B_1, r_1 and  r_2 into equation (2) to get B_2

B_2=(r_1^3/r_2^3)B_1

=0.7 * 10^-5 T

The magnet applies a magnetic field  0.7 * 10^-5 T at distance 0.51 m

A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much time does it take the marble to hit the ground? b. What was the velocity of the marble on the table?

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

Pressing two objects together with more force ______ friction. A. Increases
B. Has no effect on
C. Removes
D. Reduces
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