a. The magnitude of the acceleration when the object reaches maximum height is 9.8 m/s² and it is directed downwards
b. The maximum vertical height of the object is 44.55 m
c. The horizontal range R is 134.3m
The question has to do with projectile
<h3>What is a projectile?</h3>
A projectile is an object thrown into the air that follow a parabolic path
<h3>a. What is the magnitude and direction of the acceleration when the object reaches the maximum height?</h3>
At maximum height, the only acceleration on the object is acceleration due to gravity which is 9.8 m/s² and it is directed downwards.
So, magnitude of the acceleration when the object reaches maximum height is 9.8 m/s² and it is directed downwards
<h3>b. What is the maximum vertical height of the object?</h3>
The maximum height of the object is given by h = u²sin²Ф/g where
- u = initial speed of the object = 37.0 m/s
- Ф = angle of projection = 53° and
- g = acceleration due to gravity = 9.8 m/s²
Substitiuting the values of the variabloes intot he equation, we have
h = u²sin²Ф/2g
h = (37.0 m/s)²sin²53°/(2 × 9.8 m/s²)
h = 1369 m²/s² × (0.0.7986)²/19.6 m/s²
h = 1369 m²/s² × 0.6378/19.6 m/s²
h = 873.174 m²/s²/19.6 m/s²
h = 44.55 m
So, the maximum vertical height of the object is 44.55 m
<h3>c. What is the horizontal range R</h3>
The horizontal range R is given by R = u²sin2Ф/g where
- u = initial speed of the object = 37.0 m/s
- Ф = angle of projection = 53° and
- g = acceleration due to gravity = 9.8 m/s²
Substitiuting the values of the variabloes intot he equation, we have
h = u²sin2Ф/2g
h = (37.0 m/s)²sin2(53°)/9.8 m/s²
h = 1369 m²/s² × sin106°/9.8 m/s²
h = 1369 m²/s² × 0.9613/9.8 m/s²
h = 1315.97 m²/s²/9.8 m/s²
h = 134.28 m
h ≅ 134.3 m
So, the horizontal range R is 134.3m
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