Question

a) 43
12. A person can throw a stone to a maximum
distance of 100m. The greatest height to which
he can throw the stone is:
a) 100m
b) 75m c) 50m d) 25m​

Answer 1

Answer:

Step-by-step explanation:

The total range for a projectile: R = ((Vo^2)sin2θ)/g

The greatest range is achieved when using a 45° angle:

Vo^2 = gR/sin2θ = (9.8*100)/sin(2*45) = 980/sin90 = 980

Then use the following formula: V^2 = Vo^2 + 2ah.

The greatest height is achieved when thrown straight up at 90° angle.

When thrown vertically upwards at max height V^2 has reduced to zero

h = (V^2-Vo^2)/2a = (0^2-980)/(2*-9.8) = -980/-19.6 = 50 m

Note: This is supposed to be calculated with the angle but since sin90° = 1 you

can skip it in the calculation since the vertical component of initial velocity:

(Voy)^2 = (Vosinθ)^2 = (Vo)^2(sinθ)^2 = 980(sin90)^2 = 980*1 = 980

So the greatest height is 50 m, when thrown at 90° angle. 25 m is the greatest

height when thrown at 45 °