Answer:

PROJECTILE

A baseball player leads off the game and hits dong home run. The ball leaves the bat at an angle of 45° with a velocity of 20 m/s.

Given:

v¡ = 20 m/s

θ = 45°

g = 9.8m/s²

A. What is the maximum height reached by the ball?

d_{y} = \displaystyle\frac{(v_{i} \sin \theta)^{2}}{2g}d

y

=

2g

(v

i

sinθ)

2

d_{y} = \displaystyle\frac{(20 \: m/s \times \sin 45^{\circ})^{2}}{2(9.8 m/s^{2})}d

y

=

2(9.8m/s

2

)

(20m/s×sin45

∘

)

2

d_{y} = 10.20 \: md

y

=10.20m

Answer: \: \: \sf \large \bold{ \blue{10.20 \: m}}10.20m

B. What is the horizontal displacement (range )of the ball?

R = \displaystyle\frac{v_{i} \: ^{2} \sin2 \theta}{g}R=

g

v

i

2

sin2θ

R = \displaystyle\frac{(20 \: m/s)^{2} \sin 90^{\circ}}{9.8 m/s^{2}}R=

9.8m/s

2

(20m/s)

2

sin90

∘

R = 40.82 \: mR=40.82m

Answer: \: \: \sf \large \bold{ \blue{40.82 \: m}}40.82m

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