Question

A ball is projected horizontally with speed u from height 55m above the ground. If the ball just hits the target at height 10.9m above the ground and is at horizontal distance 180m. Find u.
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Answer 1

Answer:

u can also refer to the attached file

Let the initial vel of ball be v and time taken for its light be t 

Then, 

Also, in vertical displacement

(10.9−55)=(0)t−2gt2  [initial vertical component is zero]

⇒−−9.844.1×2=t

⇒3s=t

From (i) and (ii)

v(3)=180

v=60 m/s

Answer 2

Answer:

60.67 m/s

Explanation:

initial velocity = u in the horizontal direction. The projectile travels with the constant velocity u in the horizontal direction. g = 10 m/s^2.

Time taken to travel a horizontal distance of 180 m

T = 180/u sec.,

Height dropped by the projectile in T sec is = 55 - 10.9 = 44.1 m.

For the vertical direction, apply the kinematics formula :

s = u t - 1/2 g T^2

44.1 = 0 - 5 T^2

T = √8.802 sec.

=> u = 180/T = 60.67 m/s