Question

Example 65. A body is thrown horizontally from the top of
a tower and strikes the ground after three seconds at an angle
of 45° with the horizontal. Find the height of the tower and
the speed with which the body was projected. Take
8 =9,8 ms?
solusion :−
given , t=3
tan β=45°
sy=uyt+(1/2)gt^2
sy=0t+(1/2)×(9.8)×(3)^2
sy=((9.8)/2)×9
sy=4.41m

vy=uy+gt
vy=0+(9.8)×(3)
vy=29.4m/s

⇒vx=ux
29.4=29.4m/s

tan 45°=((vy)/(vx))
=((29.4)/(29.4))
=1​

Answer 1

Explanation:

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Answer 2

Answer:

Given :        uy=0                    ay=g=9.8m/s2

Let the projectile's initial velocity be  u.

Time of flight      T=3  s             (given)

Using         T=g2h                     ⟹h=2gT2

∴   h=29.8×32=44.1 m

x direction :      ax=0             ⟹Vx=u

y direction :     Vy=uy+ayT

∴   Vy=0+9.8×3=29.4m/s

Also       VxVy=tan45o=1

⟹     Vx=Vy=29.4  m/s

Thus initial speed of the projectile    u=Vx=29.4m/s