Question

A person can throw a stone to a maximum distance of 100 m . the greatest height to which he can throw the stone is

Answer 1

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°.

Answer 2

Answer:50

Step-by-step explanation:

R=100m

R=u^2 sin2theta/g

100=u^2sin2×45/10 since it reaches the max height it will angle 45°

U^2=1000

H=u^2 sin^2theta /2g

H=(1000×sin^2 45°)/20

H=1000/20

H=50m