Question

A ball positioned at the top of a tower 90m high is thrown horizontally with a velocity of 25m/S.How far away from the from the tower on the ground does the ball hit(g=10m/s^2

Answer 1

Answer:

105 m

Explanation:

t =

$\sqrt{2h \div g}$

hence by calculation,

t =

$\sqrt{2 \times 90 \div 10 } = 3 \sqrt{2} s$

Ux ( horizontal velocity ) 25 m/s

range or horizontal distance = Ux × t

= 25 × 3 ×(2)½

=

$75 \sqrt{2}$

=

105 m