Jawaban:

1. y = 1,21 m

2. Rₘₐₓ = 22,5 m

Penjelasan:

Diketahui :

θ = 45°

v₀ = 15 m/s

g = 10 m/s²

Ditanya :

1. y saat t = 2 s

2. Rₘₐₓ

Jawab :

1. y saat t = 2s

y = v₀ᵧ.t - ½.g.t²

y = (v₀ sin θ).t - ½.g.t²

y = ( 15.sin 45° m/s ).(2 s) - ½.(10 m/s²).(2 s)²

y = ( 15 × ½√2 m/s ).(2 s) - 20 m

y = 15√2 m - 20 m

y = 21,21 - 20 m

y = 1,21 m

2. jarak maksimum bola tercapai saat y = 0

y = v₀ᵧ.t - ½.g.t²

y = (v₀ sin θ).t - ½.g.t²

y = ( 15.sin 45° ).t - ½.(10).t²

0 = ( 15 × ½√2 ).t - 5t²

0 = 15⁄2√2 t - 5t²

0 = 5t.(3⁄2√2 - t)

5t = 0 atau 3⁄2√2 - t = 0

t = 0 tₘₐₓ = 3⁄2√2 s

tₘₐₓ = 2,12 s

jarak maksimum tercapai saat tₘₐₓ = 3⁄2√2 s

Rₘₐₓ = vₒₓ.tₘₐₓ

= vₒ cos θ . tₘₐₓ

= 15.sin 45° . 3⁄2√2

= 15 × ½√2 × 3⁄2√2

= 15 × 3⁄2

= 45⁄2

= 22,5 m