Question : A ball is thrown at an angle θ and another ball is thrown at an angle (90-θ) with the horizontal direction from the same point with velocity 39.2 m/s. The second ball reaches 50 m higher than the first ball. Find their individual heights. (g=9.8 m/s²).
Doubt by Aakarsh
Solution :
u=39.2 m/s
H2=50+H1
Case I
u=39.2 m/s
θ1=θ
H1=u2sin2θ1/2g
H1=u2sin2θ/2g — (1)
H1=u2sin2θ/2g — (1)
Case 2
u=39.2 m/s
θ2=(90-θ)
H2=u2sin2θ2/2g
H2=u2sin2(90-θ)/2g
H2=u2cos2θ/2g — (2)
H2=u2sin2(90-θ)/2g
H2=u2cos2θ/2g — (2)
Adding equation (1) and (2)
H1+H2=u2sin2θ/2g + u2cos2θ/2g
H1+50+H1=u2[sin2θ+cos2θ]/2g [∵H2=50+H1]
2H1+50=u2[sin2θ+cos2θ]/2g
2H1+50=u2/2g
2H1+50=(39.2)²/2(9.8)
2H1+50=78.4
2H1=78.4-50
2H1=28.4
H1=28.4/2
H1=14.2 m
H2=50+H1
H2=50+14.2
H2=64.2 m
H2=64.2 m
Hence, their individual heights are 14.2 m and 64.2 m.
Similar Question :
A ball is thrown at an angle θ and another ball is thrown at angle (90-θ) with the horizontal direction from the same point each with speed of 40 m/s . The second ball reach 50 m higher than the first ball . Find their individual heights. Take g=10 m/s.
Ans : 15 m and 65 m