Question

A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45o45o . With what speed was it thrown?

Answer 1

Answer:

V = 19.79 m/s

Explanation:

given,

height of the hill, h = 20 m

angle of strike,θ = 45°

Speed of thrown = ?

using equation of motion for time calculation

$s = ut+\dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 20}{9.8}}$

  t = 2.02 s

Horizontal velocity = v_x = V

Vertical velocity = v_y= g t

                              v_y = 9.8 x 2.02 = 19.79 m/s

now, we know,

$tan \theta = \dfrac{v_y}{v_x}$

$tan 45^0 = \dfrac{19.79}{V}$

  V = 19.79 m/s

Hence, the speed at which ball was thrown is equal to 19.79 m/s.

Answer 2

The speed will be "19.79 m/s".

Given:

Height of the hill,

• h = 20 m

Angle of strike,

• $\Theta$ = 45°

By using the equation of motion, we get

→ $s = ut +\frac{1}{2} gt^2$

or,

→ $t = \sqrt{\frac{2s}{g} }$

By substituting the values, we get

→    $= \sqrt{\frac{2\times 20}{9.8} }$

→    $= 2.02 \ s$

Now,

Horizontal velocity,

• $v_x = V$

Vertical velocity,

• $v_y = gt$

            $= 9.8\times 2.02$

            $= 19.79 \ m/s$    

As we know,

→ $tan \Theta = \frac{v_y}{v_x}$

$tan 45^{\circ} = \frac{19.79}{V}$

        $V = 19.79 \ m/s$

Thus the response above is correct.

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