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A rock is kicked horizontally at15m/sfrom a hill with a45oslope (Fig. 3-58). How long does it take for the rock to hit the ground?

FIGURE 3-58Problem 72

Short Answer

Expert verified

The rock hits the ground in 3.06 seconds after being launched.

Step by step solution

01

Step 1. Time of flight

The time of flight refers to the time period for which a projectile remains in the air.

02

Step 2. Given data and assumptions

The initial velocity of the rock,v0=15m/s

Let us assume that at the point of landing, the horizontal displacement is x, the vertical displacement is y, and the time taken by the rock to hit the slope is t.

03

Step 3. Setting up equations for horizontal motion

The initial velocity of the rock in the horizontal direction,vx=v0

Using the second equation of motion for the horizontal motion, you can write:

x=vxt+0x=v0t

(As acceleration in the horizontal direction is zero)

Substituting the values in the above equation, you get:

role="math" localid="1644994977727" x=15t(i)

04

Step 4. Setting up equations for vertical motion

The initial velocity along the vertical direction will be:

vy=0(The projectile has been launched horizontally)

Applying the second equation of motion for the vertical direction, you get:

y=0+12gt2

Substituting the values in the above equation, you get:

y=0+12×-9.8×t2

The above equation can be rewritten as:

role="math" localid="1644995045620" y=-4.9t2(ii)

05

Step 5. Calculating the value of time taken

As the angle of slope is 45°, and the value of tan45°=1, the magnitude of displacement in the vertical direction is equal to the displacement in the horizontal direction.

x=y

Substituting the values from equations (i) and (ii), you get:

15t=4.9t2

Solving the above equation for t (ignoring solution t = 0 s), you get:

t = 3.06 s

Thus, the rock will hit the slope at time t = 3.06 s.

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FIGURE 3-44 Problem 43

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