Question

A ball is thrown horizontally off a cliff. If the initial speed of the ball is (15.0 + A) m/s and the cliff is (25.0 + B) m high, how far from the base of the cliff will the ball land in the water below? Calculate the answer in meters (m) and round to three significant figures.A=4B=54

Answer 1

Answer:

76.3 m

Explanation:

We are given that

Initial speed of the ball,u=(15+A)m/s

Height of cliff,h=(25.9+B) m

We have to find the distance from the base of the cliff the ball will land in the water below.

A=4 and B=54

Distance=$u\sqrt{\frac{2h}{g}}$

Using the formula and substitute the values

$D=(15+4)\sqrt{\frac{2(25+54)}{9.8}}$

Because $g=9.8m/s^2$

$D=19\sqrt{\frac{158}{9.8}}$

D=76.3

Hence, the distance from the base of the cliff the ball will land in the water below=76.3 m