Question

A rock is thrown horizontally from the top of a cliff 98 m high, with a horizontal speed of 27
m/s.
(a) For what interval of time is the rock in the air?
(b) How far from the base of the cliff does the rock land?
(c) With what velocity does the rock hit?

Answer 1

Answer:

Horizontal velocity doesn't determine the time. Vertical velocity was zero. So it's same as time taken by a stone dropped from the height.

50 = 1/2*g*t^2. Assuming g =10. t=3.16 sec

How far from the bridge? 25m/s * 3.16 = 79m

How far from the point you stand sqrt(79^2+50^2) = 93m

Answer 2

Given: A rock is thrown horizontally from the top of a cliff 98 m high, with a horizontal speed of 27 m/s.

To find:

(a) Interval of time for which the rock was in the air.

(b) Distance at which the rock lands from the base of the cliff.

(c) Velocity of the rock with which it hits.

Solution:

(a)

• The time of flight of the rock or the time interval for which the rock was in the air, is calculated using the formula,

       $t = \sqrt{\frac{2h}{g} }$

• Here, t is the time of flight, h is the height of the cliff and g is the acceleration due to gravity.
• The value of acceleration due to gravity is 9.8 ms⁻².

       $t = \sqrt{\frac{2*98m}{9.8ms^{-2} } }$

          $= 2\sqrt{5} s$

(b)

• The distance at which the rock lands from the base of the cliff or the horizontal distance is calculated as,

        $x = ut$

• Here, x is the horizontal distance, u is the initial velocity and t is the time of flight.

        $x = (27ms^{-1}) (2\sqrt{5} )$

           $= 54\sqrt{5} m$

(c)

• The velocity of the rock with which it hits the ground is calculated as,

        $v = \sqrt{u^{2} +g^{2} t^{2} }$

• On replacing the terms with the values given in the question,

        $v = \sqrt{(27)^{2} + (9.8)^{2} (2\sqrt{5})^{2} }$

           $= 51.48 ms^{-1}$

Therefore,

(a) Interval of time for which the rock was in the air is 2√5 s.

(b) Distance at which the rock lands from the base of the cliff is 54√5 m.

(c) Velocity of the rock with which it hits the ground is 51.48 ms⁻¹.