1- A rock is tossed straight up with a speed of 15.7 m/s. When it returns, it fa

Question

1- A rock is tossed straight up with a speed of 15.7 m/s. When it returns, it falls into a hole 14.4 m deep. How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

2- A football is kicked straight up into the air; it hits the ground 4.3 s later. What was the greatest height reached by the ball? Assume it is kicked from ground level.

3- A driver has a reaction time of 0.4 s, and the maximum deceleration of her car is 5.6 m/s2. She is driving at 18 m/s when suddenly she sees an obstacle in the road 50 m in front of her. How much distance is between the obstacle and her when she comes to a stop?

4- When striking, the pike, a predatory fish, can accelerate from rest to a speed of 3.9 m/s in 0.11 s. How far does the pike move during this strike?

5- In an 10.31 km race, one runner runs at a steady 10.9 km/h and another runs at 13.6 km/h. How far from the finish line, in km, is the slower runner when the faster runner finishes the race?

6- A car can go from 0 to 60 mph in 12 s. A second car is capable of twice the acceleration of the first car. Assuming that it could maintain the same acceleration at higher speeds, how much t ime will this second car take to go from 0 to 120 mph?

Explanation / Answer

problem 1)

final velocity v^2 = u^2 + 2gs

or, v^2 = 15.7^2 + 2x14.4x9.81

or, v = 23.0 m/s

and now, the rock in the air, from the instant it is released until it hits the bottom of the hole will take t time

v = u - gt

or, 23.0 = 15.7 - 9.81t

or, t = 0.744 second ................................................................ANS

PROBLEM 2:

Use the standard trajectory equations:
Time of flight = 2Vyi/g where Vyi is the initial vertical velocity
Max height = Vyi^2/2*g
4.3 = 2*Vyi/9.81 => Vyi = 21.09m/s
Max Height = Vyi^2/2g = 21.09^2/2x9.81 = 22.67m ...................................ANS

PROBLEM 3:

The reaction time tells us how far the driver goes before she starts to slow down.

d = vt
d = 18 m/s * 0.4 s = 7.2 meters

So she only has 50 m - 7.2 m = 42.8 m to stop

So now we do accelerated motion

d = 0.5 *a * t^2

Now, velocity v = at

v = 18 m/s
a = 5.6 m/s^2

or, 18 m/s = 5.6 m/s^2 * t

or, t = 20/6 = 3.21 s

so, required d = 0.5 * 5.6 * 3.21^2 = 28.85 meters .......................................ANS

PROBLEM 4:

if the acceleration is constant, what is the acceleration of the pike.

which is just 3.9/0.11 = 35.45m/s/s

how far it moves is a simple kinematic equation

s = ut + 1/2at^2
u = initial velocity = 0
s = (1/2)x35.45x0.11^2 = 0.214m ...................................................ANS

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