Question : A ball is kicked at an angle of 30° with the vertical if the horizontal component of its velocity is 19.6 m/s. Find the maximum height and horizontal range.
Doubt by Jaskirat
Solution :
θ= 90°-30°
θ= 60°
θ= 90°-30°
θ= 60°
Horizontal Component of velocity
ux = 19.6 m/s
ucosθ = 19.6
cos60° =19.6
u(1/2) = 19.6
u = 19.6×2
u = 39.2 m/s
ucosθ = 19.6
cos60° =19.6
u(1/2) = 19.6
u = 19.6×2
u = 39.2 m/s
Maximum Height attained by the Projectile
Hmax = u²sin²θ/2g
= (39.2)²sin²(60°)/[2×9.8]
= [39.2×39.2×(√3/2)²]/[2×9.8]
Hmax = u²sin²θ/2g
= (39.2)²sin²(60°)/[2×9.8]
= [39.2×39.2×(√3/2)²]/[2×9.8]
= [39.2×39.2×(3/4)]/[2×9.8]
= [39.2×39.2×3]/[4×2×9.8]
= 58.8 m
= [39.2×39.2×3]/[4×2×9.8]
= 58.8 m
Horizontal Range of Projectile
= u²sin2θ/g
= [(39.2)²sin(2×60°)]/9.8
= [39.2×39.2×sin120°]/9.8
= [39.2×39.2×(√3/2)]/9.8
= [39.2×39.2×√3)]/[2×9.8]
= [(39.2)²sin(2×60°)]/9.8
= [39.2×39.2×sin120°]/9.8
= [39.2×39.2×(√3/2)]/9.8
= [39.2×39.2×√3)]/[2×9.8]
= 78.4√3
= 78.4×1.732
= 135.788
= 135.79 m
= 78.4×1.732
= 135.788
= 135.79 m
Hence, maximum height is 58.8 m and Range is 135.79 m