Question

A projectile is launched with speed v0 and angle θ. Derive an expression for the projectile's maximum height h.

A baseball is hit with a speed of 35.0 m/s . Calculate its height if it is hit at an angle of 30.0 ∘ .

A baseball is hit with a speed of 35.0 m/s . Calculate its distance traveled if it is hit at an angle of 30.0 ∘ .

A baseball is hit with a speed of 35.0 m/s . Calculate its height if it is hit at an angle of 45.0 ∘ .

A baseball is hit with a speed of 35.0 m/s . Calculate its distance traveled if it is hit at an angle of 45.0 ∘ .

A baseball is hit with a speed of 35.0 m/s . Calculate its height if it is hit at an angle of 60.0 ∘ .

A baseball is hit with a speed of 35.0 m/s . Calculate its distance traveled if it is hit at an angle of 60.0 ∘ .

Answer 1

a.)

At max height vertical velocity component will be zero, So Using 3rd kinematic equation

V1y^2 = V0y^2 + 2*ay*Hmax

Hmax = (V1y^2 - V0y^2)/(2*ay)

here,

V0 = initial speed(v0) at angle with horizontal

V0x = v0*cos

V0y = v0*sin

ay = acceleration due to gravity = -g = -9.81 m/s^2

V1y = final vertical speed at maximum height = 0

So, Hmax = (0 - (v0*sin)^2)/(2*-g)

Hmax = (v0*sin)^2/(2*g)

Hmax = 0.051*(v0*sin)^2

b.)

now, v0 = 35.0 m/s

and = 30 deg

So, Hmax = (35.0*sin 30.0 deg)^2/(2*9.81)

Hmax = 15.61 m

distance travelled is given by,

Range(R) = v0^2*sin(2*)/g

So, R = 35.0^2*sin(2*30.0 deg)/9.81

R = 108.14 m

c.)

now, v0 = 35.0 m/s

and = 45 deg

So, Hmax = (35.0*sin 45.0 deg)^2/(2*9.81)

Hmax = 31.22 m

distance travelled is given by,

Range(R) = v0^2*sin(2*)/g

So, R = 35.0^2*sin(2*45.0 deg)/9.81

R = 124.87 m

d.)

now, v0 = 35.0 m/s

and = 60.0 deg

So, Hmax = (35.0*sin 60.0 deg)^2/(2*9.81)

Hmax = 46.83 m

distance travelled is given by,

Range(R) = v0^2*sin(2*)/g

So, R = 35.0^2*sin(2*60.0 deg)/9.81

R = 108.14 m

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