An athlete throws a javelin from ground level such that its initial vertical component of … — JAMB UTME

An athlete throws a javelin from ground level such that its initial vertical component of velocity is $40\text{ m/s}$ and its initial horizontal component of velocity is $30\text{ m/s}$ . What is the horizontal distance in metres covered by the javelin when it reaches a height of $35\text{ m}$ above the ground for the first time? (Take $g = 10\text{ m/s}^2$ )

Cevap: 30 m

Cevap

The horizontal distance covered by the javelin when it reaches a height of 35 m for the first time is 30 m.

Using

y = u_y t - \frac{1}{2}g t^2

with

u_y = 40\text{ m/s}

y = 35\text{ m}

, and

g = 10\text{ m/s}^2

yields

35 = 40t - 5t^2

. Dividing by 5 gives

t^2 - 8t + 7 = 0

, which factors to

(t - 1)(t - 7) = 0

. The roots are

t = 1\text{ s}

(ascent) and

t = 7\text{ s}

(descent). For the first time,

t = 1\text{ s}

. The horizontal displacement is

x = u_x t = 30\text{ m/s} \times 1\text{ s} = 30\text{ m}

Adım Adım Çözüm

Set up the vertical motion equation to find the time when height is 35 m

35 = 40t - 5t^2

Vertical displacement in projectile motion depends on the vertical initial velocity component and acceleration due to gravity.

Solve the quadratic equation for time

t

t^2 - 8t + 7 = 0 \implies t = 1\text{ s} \text{ or } t = 7\text{ s}

A projectile reaches a given non-peak height twice: once ascending and once descending.

Select the first time value and calculate horizontal distance

x = u_x \times t = 30 \times 1 = 30\text{ m}

Horizontal velocity remains constant throughout the flight, so distance is speed multiplied by time.

Anahtar Kavram

Independence of vertical and horizontal components in projectile motion

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru