Question

a projectile is launched from ground level to the top of a cliff which is 95 m away and 155 m high. If the projectile lands on top of the cliff 7.6 s after it is fired, what was its initial velocity? (Be careful what you assume).

Answer 1

Answer:

Initial velocity ,v =24.6 m/s with direction 58.5⁰ from positive x axis.

Explanation:

Given,

Height of cliff =155 m

Distance to cliff = 95 m

Time for projectile to reach top of cliff =7.6 s

See the figure below

AB is the cliff. C is the point from where the projectile is launched .

Let the initial velocity of the projectile be V.The angle made by V with the x axis is .

V has two components:

1) V_x=v cos in the horrizontal direction.

2)Vy=v sin in the vertical direction.

We know that distance = velocity *time or D =Vt

In the horrizontal direction,

D_x=V_xt -------------(1)

In the verical direction,

D_y = V_yt -------------(2)

From figure,

The horrizontal distance travelled by projectile,D_x =95 m

The verical distance travelled by projectile , Dy= 155 m

Also we know that time taken, t =7.6 s

Substituting the given values i equation (1) and (2),

From (1) ,

95 =V_x(7.6)  

95=vcos (7.6)

v cos =95/7.6=12.5 ---------(3)

From (2),

155=V_y(7.6)

155= v sin (7.6)

v sin=155/7.6=20.39 ----(4)

Divide equation (3) and (4),

So tan =1.63

=tan^-1(1.63)

=58.5⁰

Substitute value of in equation (4) ,

v sin(58.5) =20.39

v (0.85) =20.39

v=20.39/0.85 =24.6

There fore initial velocity ,v =24.6 m/s with direction 58.5⁰ from positive x axis.