Question

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Constants Part A A man stands on the roof of a building of height 14.3 m and throws a rock with a velocity of magnitude 31.1 m/s at an angle of 33.6 above the horizontal. You can ignore air resistance Calculate the maximum height above the roof reached by the rock. A2op SubmitP Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining Part B Calculate the magnitude of the velocity of the rock just before it strikes the ground 0 m/s Submit Request Answer

Answer 1

Given :

y₀ = 14.3 m , x₀ = 0 m , v₀ = 31.1 m/s ,$\theta$ = 33.6° , g = 9.81 m/s²

x components :

v_0x = v₀cos(33.6) = (31.1)cos(33.6) = 25.90 m/s

y components :

v_0y = v₀sin(33.6) = (31.1)sin(33.6) = 17.21 m/s

Solution :

(a) maximum height above the roof reached by the rock

Using equation of motion,

v_yf² = v_0y² + 2ah

Here , a = -g  

And velocity at the maximum height would be zero, i.e. v_yf =0

So,

0 = (17.21)² -2(9.81)h

h = (17.21)²/19.62 = 15.1 m

Answer : h = 15.1 m

(b) Magnitude of velocity of rock just before it strikes the ground

The horizontal velocity will be constant , so v_fx = v_0x = 25.90 m/s

The vertical velocity ia given by the equation of motion as :

v_yf² = v_0y² + 2g(h+14.3)

v_yf² = 2(9.81)(15.1+14.3) = 576.83

v_yf = 24.02 m/s

The magnitude of velocity is given by:

v_f = (v_fx² + v_fy²)^1/2

= (25.90²+24.02²)^1/2

= 35.32 m/s

Answer : 35.32 m/s

(c) Horizontal range

Time required to reach at ground is given by:

y = v_0yt + (1/2)at²

-14.3 = (17.21)t + (1/2)(-9.81)t²

-4.905 t² + 17.21t + 14.3 =0

By solving we get , t = 4.202 , - 0.694

Time can not be negative , so t = 4.202 s

The horizontal range is given by:

x = v_0xt+ (1/2)at²

x = (25.90)(4.202) = 108.83 m

Answer : x = 108.83 m