The height [tex]y[/tex] and distance [tex]x[/tex] along the horizontal for a body projected in the vertical plane are given by [tex]y=8t-5t^2[/tex] and [tex]x=6t[/tex]. The initial velocity of the projected body is

(a) [tex]8 \, m/s[/tex]
(b) [tex]9 \, m/s[/tex]
(c) [tex]10 \, m/s[/tex]
(d) [tex]\frac{10}{3} \, m/s[/tex]



Answer :

Let's solve the problem step-by-step.

1. Given equations:
- The height of the body as a function of time [tex]\( t \)[/tex] is given by [tex]\( y = 8t - 5t^2 \)[/tex].
- The horizontal distance of the body as a function of time [tex]\( t \)[/tex] is given by [tex]\( x = 6t \)[/tex].

2. Determine the initial vertical velocity component [tex]\( v_{y0} \)[/tex]:
- The height equation [tex]\( y = 8t - 5t^2 \)[/tex] represents a quadratic equation, where the coefficient of [tex]\( t \)[/tex] in the linear term (8) is the initial vertical velocity.
- Therefore, [tex]\( v_{y0} = 8 \, \text{m/s} \)[/tex].

3. Determine the initial horizontal velocity component [tex]\( v_{x0} \)[/tex]:
- The horizontal distance equation [tex]\( x = 6t \)[/tex] represents a linear equation, where the coefficient of [tex]\( t \)[/tex] (6) is the initial horizontal velocity.
- Therefore, [tex]\( v_{x0} = 6 \, \text{m/s} \)[/tex].

4. Calculate the resultant initial velocity [tex]\( v_0 \)[/tex]:
To find the magnitude of the initial velocity, we use the Pythagorean theorem. The resultant velocity is the vector sum of the horizontal and vertical components.

[tex]\[ v_0 = \sqrt{v_{x0}^2 + v_{y0}^2} \][/tex]

Substituting the given values:

[tex]\[ v_0 = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s} \][/tex]

Thus, the initial velocity of the body is [tex]\( \boxed{10 \, \text{m/s}} \)[/tex].

5. Select the correct option:
From the choices provided, the correct answer is:
[tex]\[ (c) \, 10 \, \text{m/s} \][/tex]