A ball is projected at an angle of 60° with the verticle from the top of a tower whose height is 50m from the ground. If the speed of the ball is 20m/s then find H achieved by the ball from the ground and the R from the bottom of the tower.
Answers
Answer:
Height from ground = 25 m
Range from bottom = 56 m
Explanation:
A ball is projected at an angle of 60° with the verticle from the top of a tower whose height is 50m from the ground. If the speed of the ball is 20m/s then find H achieved by the ball from the ground and the R from the bottom of the tower.
A ball is projected at an angle of 60° with the vertical from the top of a tower at 20m/s
=> Vertical Velocity upwards = 20cos60 = 10 m/s
At top Velocity = 0
a = -g = -10 m/s²
Using V = u + at
=> 0 = 10 -10t
=> t = 1 sec
Using S = ut + (1/2)at²
Vertical Distance covered in 1 sec = 10*1 + (1/2)(-10)*1² = 5 m
Vertical Height achieved from ground = 20 + 5 = 25 m
Horizontal Speed = 20Sin60 = 10√3
Horizontal Distance Covered in 1sec = 10√3*1 = 10√3 m
Time further taken to ball to reach at ground from 25 m
25 = (1/2)(10)t²
=> t² = 5 sec
=> t = √5
Horizontal Distance Covered in next √5 sec = 10√3*√5 = 10√15
Range from bottom of Tower = 10√3 + 10√15 = 56 m
Height from ground = 25 m
Range from bottom = 56 m