Sample Problems Chapter 3 (Vectors and Two-Dimensional Motion) 1. A hiker stands

Question

Sample Problems Chapter 3 (Vectors and Two-Dimensional Motion) 1. A hiker stands at the edge of a hilltop and throws a rock horizontally over the edge with an initial speed of 13.0 m/s. (Take the height of the rock's release from the hill to be 61.0 m.) If g = 9.8 m/s2 and air resistance is negligible, then a) how long does it take the rock to hit the ground below? b) Determine the speed and impact angle of the rock the instant betore it strikes the ground. 2. A projectile is launched with an initial speed of 75 m/s at an angle of 25 above the horizontal. The projectile lands on a hillside 6.5 s later. If air resistance is neglected, a) what is the projectile's velocity at the highest point of its trajectory? b) What is the linear distance from where the projectile was launched to where it hits its target? 3. A car travels due east with a speed of 56.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth. The traces of the rain on the side windows of the car make an angle of 70.0 with the vertical. Find the velocity of the rain with respect to (a) the earth and (b) the car

Explanation / Answer

1.) Vertical distance to be covered = 61 m

vertically, S = ut + 0.5 gt2

So, 61 = 0t + 0.5 x 9.8 x t2

61 / (0.5 x 9.8) = t2

t = 3.528311 seconds

b.) When it hits the ground, the vertical component of velocity be Vv  

Vv2 - 02 = 2 x 9.8 x 61

Vv =34.5774493 m/s

And horizontal component of velocity Vh remains constant throughout

Speed = ( Vv2 + Vh2 )1/2 = ( 34.57744932 + 132 )1/2 = 36.94 m/s

So, impact angle = tan-1 (Vv / Vh ) = tan-1 ( 34.5774493 / 13 ) = 69.3953 degrees

2.) At the highest point the vertical component of the velocity becomes zero and the total velocity is nothing but the Horizontal component of the velocity which will be Vh = V Cos = 75 Cos25 = 67.97308 m/s

Time of flight = 2VSin/g = 2 x 75 x Sin 25 / 9.8 = 6.4686 seconds ~ 6.5 seconds

So, after the 6.5 seconds, the linear distance covered is nothing but the range of the projectile which is given by

R = V2 Sin2 /g = 75 x 75 x Sin 50 / 9.8 = 439.6938 m

3.) Let the velocity of the rain be Vv vertically downwards

with respect to the car, the velocity of the rain will be Vv downwards + Vcar horizontal

Since the angle made with the vertical is 70 degrees, it means

tan 70 = Vcar / Vv = 56 / Vv

Vv = 56 / tan 70 = 20.38233 km /hr (answer to a)

with respect to the car, the speed will be V = ( Vv2 + Vcar2 )1/2 = ( 20.382332 + 562 ) = 59.59 kmph (answer to b)

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