Help me to do this PHYSICS question, please????????? Thank You :)?

Question

A projectile is shot from the edge of a cliff h = 305 m above ground level with an initial speed of v0 = 105 m/s at an angle of 37.0° with the horizontal.

(a) Determine the time taken by the projectile to hit point P at ground level= .... s

(b) Determine the range X of the projectile as measured from the base of the cliff = .... m

(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.)
horizontal= ... m/s
Vertical= .... m/s

(d) What is the the magnitude of the velocity?.... m/s

(e) What is the angle made by the velocity vector with the horizontal?

(f) Find the maximum height above the cliff top reached by the projectile.
...... m

Answer 1

(a) Vertical:(work with the vertical to get the time of flight)
v(zero y) = 105sin37 = 63.1906m/s
We'll use the distance formula which will give us a quadratic in t.
s = s(zero) + v(zero)t + (1/2)at^2
s = 305 + 63.1906t + (1/2)(-9.8)t^2
4.9t^2 - 63.1906 - 305 = 0
t = (-B +/- √[B^2 - 4AC])/2a
t = (63.1906 +/- √[-63.1906^2 - 4(4.9)(-305)])/9.8
t = (63.1906 +/- 99.8552)/9.8
The positive answer makes sense
t = 16.6373sec

(b) D =rt = v(zero x)t = (105cos37)(16.6373) = 1395.15m

(c)Vertical
v(final y) = v(zero) + at = 63.1906 -9.8(16.6373) = -99.855m/s
Horizontal (v(x) assumed constant)
v(final x) = 105cos37 = 83.8567m/s

(d) Magnitude = √(-99.855^2 + 83.8567^2) = 130.395m/s

(e) Angle = arctan v(y)/v(x) = arctan(99.855/83.8567) = 49.977degrees

(f) Max Height:
Vertical distance formula:
s = 305 + 63.1906t - 4.9t^2
Take the derivative and set it equal to zero:
ds/dt = 63.1906 - 9.8t = 0
9.8 t = 63.1906
t = 6.44802s
Plug back into the distance formula:
s(max) = 305 + 63.1906(6.44802) - 4.9(6.44802)^2 = 508.727m

Answer 2

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