Question

a canon ball is projected horizontally from the top of a tower 20 m high with a velocity of 70 m/s .find the time of flight, the distance from the foot of the tower to the point at which it hits the ground.

Answer 1

Answer:

the distance from the foot of the tower to point at which it hits the ground is

X0=0, X=?, Ux=70m/sec , Ax=0, t=10s

X=0+70×10+1/2×0×10^2= 700m

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Height of tower(H) = 20 m.
• Initial velocity (u) = 70 m/s.
• Acceleration due to gravity (g) = 10 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

As the ball is projected horizontally the vertical component of velocity will be Zero and the horizontal component will not be Zero.

I.e.

$\large{ u_y = 0 \: m/s}$

and,

$\large{ u_x = u = 70 \: m/s}$

Time of Flight:-

Applying second kinematic equation,

$\large{\bold{S = ut + \dfrac{1}{2} at^2}}$

Taking Vertical motion

$\large{\bold{H = u_yt + \dfrac{1}{2} a_yt^2}}$

Substituting the values,

$\large{20 = 0 + \dfrac{1}{2} \times 10 \times t^2}$

$\large{20 \times 2 = 10 \times t^2}$

$\large{ t^2 = \dfrac{40}{10}}$

$\large{t^2 = 4}$

$\large{t = \sqrt{4}}$

$\huge{\boxed{\boxed{t = 2 \: Seconds}}}$

Horizontal distance (Range) :-

Applying second kinematic equation,

$\large{\bold{S = ut + \dfrac{1}{2} at^2}}$

Taking Horizontal motion

$\large{\bold{R = u_xt + \dfrac{1}{2} a_xt^2}}$

Note:-

${a_y = 0}$

As there is no acceleration in horizontal direction,

Substituting the values,

$\large{R = u_xt +0}$

$\large{R = u_xt}$

$\large{R = 70 \times 2}$

$\huge{\boxed{\boxed{R = 140 \: meters}}}$

So, the Time taken is 2 seconds and Horizontal distance is 140 meters.