Respuesta :
Answer:
Your Answer
Explanation:
(a) At maximum height, i.e. at point P:
Vertical component of velocity v_y =0v
y
=0
Initial horizontal speed
x_{0x}=v_0cosθ \\ = 30 \times cos 33º \\ = 25.2 \;m/sx
0x
=v
0
cosθ
=30×cos33º
=25.2m/s
Initial vertical speed
v_{0y}=v_0sinθ \\ = 30 \times sin 33º \\ = 16.3 \;m/sv
0y
=v
0
sinθ
=30×sin33º
=16.3m/s
Let H be the maximum height reached by the rock, as measured from the roof.
v^2_y=v^2_{0y}-2gHv
y
2
=v
0y
2
−2gH
Maximum height
H= \frac{v^2_{0y}}{2g} \\ = \frac{16.3}{2 \times 9.8} \\ = 13.6 \;mH=
2g
v
0y
2
=
2×9.8
16.3
=13.6m
(b) Magnitude of velocity of the rock just before it strikes the ground
v = \sqrt{v^2_x+v^2_y} \\ v_x = v_{0x} \\ = 25.2 \;m/sv=
v
x
2
+v
y
2
v
x
=v
0x
=25.2m/s
Time of flight (t) is given by
-h = v_{0y}t- \frac{1}{2}gt^2 \\ 4.9t^2 -16.3t -15 =0 \\ t = \frac{16.3± \sqrt{16.3^2 + 4 \times 4.9 \times 15}}{9.8} \\ = \frac{16.3±23.7}{9.8}−h=v
0y
t−
2
1
gt
2
4.9t
2
−16.3t−15=0
t=
9.8
16.3±
16.3
2
+4×4.9×15
=
9.8
16.3±23.7
= 4.08 s (ignoring the negative value)
v_y = 16.3 – 9.8 \times 4.08 \\ = -23.7 \;m/s \\ v = \sqrt{v^2_x + v^2_y} \\ = \sqrt{25.2^2 + (-23.7)^2} \\ = 34.6 \;m/sv
y
=16.3–9.8×4.08
=−23.7m/s
v=
v
x
2
+v
y
2
=
25.2
2
+(−23.7)
2
=34.6m/s
(c) Required horizontal distance (x) = v_{0x}t(x)=v
0x
t
= 25.2 \times 4.08 \\ = 102.8 \;m=25.2×4.08
=102.8m
