Question

A projectile is fired at time t = 0.0s, from point 0 at the edge of a cliff, with initial velocity components Vox = 80 m/s and and Voy = 800 m/s. The projectile rises, then falls into the sea at point P. The time of flight of the projectile is 200.0 s

In Figure 3.2c, the height H of the cliff is closest to:

A)32,130m

B)36,000m

C)47,490m

D)43,650m

E)39,810m

Answer 1

Voy = 800m/s

time after which it will reach the level from which it was thrown is give by : s = ut+1/2at^2

here u = 800m/s

s=0

a = -9.8m/s2

thus t = 163.3s

velocity at that time = u+at = 200 - 9.8*163.3 = -800 (i.e is downward direction)

that will be treated as u for this next section.

time left in flight = 200- 163.3 = 36.7 s

thus heigh = H = ut+1/2at^2 = 35959.8 m

thus answer will be B) 36000 m

Answer 2

so we need to find H2

T=200sec

V0y=800m/sec

substituting these in the second equation we get H2=

400000-320000+2H2/10=0

80000+H2/5=0

H2=400000 metre or 400kilometres ~ 39,810 metres answer is E

Answer 3

Voy=800 m/s

g(acceleartion due to gravity)=9.81 m/s^2

after the projectile is fired, it reaches a maximum height and then it reaches a point at which its height above the ground is same as the height of the cliff above the ground. Let the latter point be C.

time taken to reach maximum height=Voy/g

time taken to reach a point which is at the same vertical height as the edge of the cliff i.e. C=2*Voy/g

=2*800/9.81

=163.1 s

time of flight of the projectile=200 s

therefore, remaining time is taken to travel the vertical distance which is same as the height of the cliff.

remaining time=200-163.1

=36.9 s

let the height of the cliff above the ground=H

after reaching point C, direction of vertical component of velocity changes. Initially it was in the upward direction but now it is in the downward direction.

now, Voy and g are in the same direction, and we can use the equation below to calculate height H.

then H= Voy*36.9+0.5*g*(36.9)^2

from here, we get H is closest to 36000 m.

Answer 4

The initial velocity of the projectile is

The time of flight is

When the projectile is projected from a edge of the cliff, then the height of the cliff is given by kinematic equations.

               

                

                 

                 

Therefore, the height of the cliff is 36000 m. The correct option is (B).

Answer 5