Question

A shot put is thrown upward with a velocity of 35 ft./sec. at a height of 4 ft. and an angle of 40°. How long will it take for the shot put to be a horizontal distance of 40 ft. from the person throwing it? Acceleration due to gravity is 32 ft./s^2

Answer 1

The horizontal component of the equation of motion is given by:


`x(t)=vt\cos\theta` . . . (1)

where: v is the velocity of projection, t is the time and theta is the angle of projection

The vertical component of the equation of motion is given by:


`y(t)=y_0+vt\sin\theta- (1)/(2)t^2` . . . (2)

where, y_0 is the height above the ground of projection.

Given that a shot put is thrown upward with a velocity of 35 ft./sec. at a height of 4 ft. and an angle of 40°.

This meanns that v = 35 ft/sec; y_0 = 4 ft; and theta = 40°.

Let the height of the shot put above the ground at the time the shot put's horizontal distance from the point of projection is 40 ft be h, then we have:


`h=y_0+vt\sin\theta- (1)/(2)t^2 \\ \\ \Rightarrow h=4+35t\sin40^o- (1)/(2) t^2 \\ \\ \Rightarrow (1)/(2) t^2-35t\sin40^o-(4-h)=0`

Solving the quadratic equation, we have:


`t= \frac{-(-35\sin40^o)\pm\sqrt{(-35)^2\sin^240^o-4*(1)/(2)*(-(4-h))}}{2\left((1)/(2)\right)} \\ \\ = (35(0.6428)\pm√(1,225(0.6428)^2+2(4-h)))/(1)=22.498\pm√(1,225(0.4132)+8-2h) \\ \\ =22.498\pm√(506.17+8-2h)=22.498\pm√(514.17-2h)`

But t must not be negative, thus we disregard the negative and thus have:


`t=22.498+√(514.17-2h)`

Substituting for t into equation (1) gives:


`x(t)=vt\cot\theta \\ \\ \Rightarrow40=35\left(22.498+√(514.17-2h)\right)\cos40^o \\ \\ =(787.43+35√(514.17-2h))(0.7660)=603.1714+26.81√(514.17-2h) \\ \\ \Rightarrow26.81√(514.17-2h)=40-603.1714=-563.1714 \\ \\ \Rightarrow√(514.17-2h)= (-563.1714)/(26.81) =-21.0060 \\ \\ \Rightarrow514.17-2h=(-21.0060)^2=441.2520 \\ \\ \Rightarrow2h=514.17-441.2520=72.918 \\ \\ \Rightarrow h= (72.918)/(2) =36.459`

Substituting for h into t, we have:


`t=22.498+√(514.17-2(36.459)) \\ \\ =22.498+√(514.17-72.918)=22.498+√(441.252) \\ \\ =22.498+21.0060=43.5`

Therefore, the time it will take for the shot put to be a horizontal distance of 40 ft. from the person throwing it is 43.5 seconds.

Answer 2

Time taken for the shot put from the person throwing it is 1.49 seconds.

Explanation:

It is given that,

Initial velocity of shot put, v = 35 ft/sec

Height, h = 4 ft

Angle,
`\theta=40^0`

horizontal distance, d = 40 ft

Acceleration due to gravity, a = 32 ft/s²

It is the case of projectile whose horizontal component is given by :

`d=vt\ \cos\theta`

`t=(d)/(v\ \cos\theta)`

`t=(40\ ft)/(35\ ft/s cos(40))`

t = 1.49 seconds

Hence, this is the required solution.