Question

A body is projected horizontally from a tower of height 50 m with velocity 20 m/s, then the time after which velocity makes an angle of 45 with the horizontal is (take g = 10 m/s2

Answer 1

Answer:

Let the height of stone is h from the group at which it makes 45° with horizontal.

horizontal initial velocity = horizontal final velocity

30 = ucos45°

u = 30/cos45° = 30√2 m/s ........(i)

for vertical motion,

v_y^2=u_y^2+2g(50-h)v

y

2

=u

y

2

+2g(50−h)

u²sin²45° = 0 + 2g(50-h) [ initially stone is thrown horizontally so, vertical initial velocity = 0]

put equation (i),

(30√2)² sin²45° = 2× 10 × (50-h)

900 × 2 × 1/2 = 20(50-h)

900 = 20(50-h)

50 - h = 45

h = 5m

hence, answer should be 5m