A ball is thrown at an angle theta and another ball is thrown at an angle (90-theta) with the horizontal direction from the same point wih velocity 39.2 m/s. The second ball reaches 50m higher than the first ball. Find their individual heights.
Answers
Answered by
61
use the formula v2 = u2 - 2aS.
For the first ball, v i.e. final velocity is 0 and u initial velocity is 39.2 Sin
Θ
m/s and a is 9.8 m/s2 and S is the distance.
1536.64 Sin2
Θ
= 19.6 S
For the second ball,
1536.64 Sin2(90-
Θ
) = 19.6 (S+50).
So I get two equations,
19.6S = 1536.64 Sin2
Θ
------- A
19.6S + 980 = 1536.64 Sin2(90 -
Θ
) -------- B
A + B
39.2S + 980 = 1536.64 Sin 90
39.2S = 1536.64 - 980 = 556.64
S = 14.2 m and S+50 is 64.2 m.
Answered by
42
height of first ball (H₁) = u²sin²Ө/2g
height of 2nd ball (H₂) =u²cos²Ө/2g
a/c to question ,
H₂ - H₁ = 50m
⇒u²cos2Ө/2g=50
⇒(39.2)² cos2Ө=50 x 19.6
⇒cos2Ө= 50/2 x 39.2 = 0.64
⇒2cos²Ө=1.64
⇒cos²Ө=0.82
sin²Ө=0.18
H₁ = (39.2)² x (0.18)/19.6
=14.4 m
H₂ = (39.2)² x (0.82) /19.6
=64.288 m
Similar questions