Let:

h metre be the height of the cliff,

s metre be the horizontal distance required,

t sec. be the time of flight for the ball thrown at an angle.

For the ball dropped vertically:

h = g 1.3^2 / 2

Using g = 9.81 m/sec^2:

h = 8.28945m …(1)

Now consider the ball thrown at an angle. The question doesn’t say whether 31deg. is the angle from the vertical or from the horizontal, or in the latter case, whether it is down or up. I am taking it to be 31deg. up from the horizontal.

The ball travels horizontally at constant speed 50cos(31) m/s, and therefore:

s = 50t cos(31) …(2)

Vertically, the ball has acceleration g downward. Its initial velocity is -50 sin(31) downward (negative as I am assuming it is thrown up).

h = -50t sin(31) + gt^2 / 2

Using h from (1):

gt^2 – [100sin(31)]t – 16.5789 = 0

t = (100sin(31) +/- sqrt(10^4 sin^2(31) + 4*16.5789*9.81)) / (2*9.81)

= (51.5038 +/- 57.4735) / (2*9.81)

Using the plus sign as we are interested in the later time of reaching the ground,

t = 5.554 sec.

Substituting this in (2):

s = 238.0m.

From the time of the dropped ball you will discover the peak of the cliff from the floor. S = 0.5 gt^2 = 4.9 x a million.3^2 = 8.281 m ======================================… next we are able to discover the time of the thrown ball to attain the floor. Taking upward direction as beneficial, The displacement of this ball is – 8.281m preliminary vertical velocity = 50 sin 31 m/s. V^2 = U^2 – 2 x 9.8 x {- 8.281} V^2 = 25.seventy 5^2 + 162.3076 V = ± 28.seventy 3 m/s while you evaluate that’s falling down we take V = – 28.seventy 3 m/s. Time t = [v-u] / g = 5.559s. ================================ in this tine the item strikes horizontally by using a distance of fifty cos 31 * 5.559s = 238.25m ======================================…