Answer:
A. Using displacement =Ut + 1/2gt²
=> 0 + 1/2 (-9.8)t²
= -4.9t²
So
h(t) = 50+ displacement
= 50 - 4.9t²
B. To reach the ground
h(t) = 0
So
50-4.9t²= 0
t = √ (50/4.9)
= 3.2s
C. Using
V = u+ gt
U= 0
V= - 9.8(3.2)
= 31.4m/s
D. If u = -9m/s
Then s = ut + 1/2gt²
5t- 1/2gt²
But distance from the ground is
=.> 50-5t- 4.8t²= 0
So t solving the quadratic equation
t= 3.58s
(a) The distance of the stone above the ground level at time t is [tex]h(t) = 50 - 4.9t^2[/tex]
(b) The time taken for the stone to strike the ground is 3.19 s.
(c) The velocity of the stone when it strikes the ground is 31.4 m/s.
(d) The time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
The given parameters;
The distance of the stone above the ground level at time t is calculated as;
[tex]h(t) = h_0 - ut - \frac{1}{2} gt^2\\\\h(t) = 50 - 0 -0.5\times 9.8t^2\\\\h(t) = 50 - 4.9t^2[/tex]
The time taken for the stone to strike the ground is calculated as;
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 50}{9.8} } \\\\t = 3.19 \ s[/tex]
The velocity of the stone when it strikes the ground is calculated as;
[tex]v =u + gt\\\\v = 0 + 3.2 \times 9.8\\\\v = 31.4 \ m/s[/tex]
The time taken for the stone to reach the ground when thrown at speed of 9 m/s is calculated as;
[tex]50 = 9t + \frac{1}{2} (9.8)t^2\\\\50 = 9t + 4.9t^2\\\\4.9t^2 + 9t - 50 = 0\\\\a = 4.9 \, \ b = 9, \ \ c = -50\\\\solve \ the \ quadratic \ equation\ using \ formula \ method\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-9 \ \ + /- \ \ \sqrt{(9)^2 - 4(4.9 \times -50)} }{2(4.9)} \\\\t = 2.41 \ s \ \ or \ \ - 4.24 \ s[/tex]
Thus, the time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
Learn more here:https://brainly.com/question/9527588
